NCERT Solutions for Class 9 Science Chapter 7 Work Energy and Simple Machines – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
Newton’s laws tell you what a force does moment by moment. This chapter gives you a shortcut: keep track of energy instead. Work done on an object equals the change in its energy — and from that one statement come the formulae for kinetic energy, potential energy, the conservation of mechanical energy, power, and the mechanical advantage of a pulley, an inclined plane and a lever.
NCERT Solutions for Class 9 Science Chapter 7 Work Energy and Simple Machines PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
NCERT Solutions for Class 9 Science Chapter 7 Work Energy and Simple Machines – Questions and Answers
All 47 questions of the chapter (textbook pages 116 – 139) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
Think It Over — Page 116Chapter opener
Think It Over
Q1What will be the magnitude of velocity of the child at the bottom of the blue slide?Chapter opening picture, page 116 — redrawn sketch. Three slides of different shapes — a blue spiral one, a straight red one and a wavy purple one — all start from the same platform at height h above the ground.
Answer
If h is the height of the platform above the ground, the child arrives at the bottom with speed v = √(2gh).
At the top: kinetic energy = 0, potential energy = mgh
At the bottom: potential energy = 0, kinetic energy = ½mv2
Conservation of mechanical energy (ignoring friction):
½mv2 = mgh
v2 = 2gh v = √(2gh)
Q2Will two children of different masses reach the bottom of the same slide with the same velocity?
Answer
Yes — both reach the bottom with the same speed, √(2gh).
½mv2 = mgh
The mass m appears on both sides and cancels
v = √(2gh) — independent of m
Q3Which of the slides will result in the largest magnitude of velocity for the child at its bottom?
Answer
Look at the picture on page 116: the blue spiral slide, the red straight slide and the purple wavy slide all begin at the same platform and end at the same ground. So the vertical drop h is the same for all three, and — ignoring friction — all three give the same speed at the bottom, v = √(2gh).
v = √(2gh) depends only on h
same h → same v, whatever the length or shape of the path
In-text Questions — Page 116Chapter opener
Q1But when forces change with time or act in complicated ways, applying these laws directly can become difficult. Is there a simpler and more powerful way to understand such situations?
Answer
Yes — keep track of energy instead of force. The work–energy theorem compares only the starting and finishing states of an object, so you never have to follow the force instant by instant.
Newton’s laws need the force at every moment: F = ma gives a, then the kinematic equations give v and s. If F keeps changing, a keeps changing and the kinematic equations no longer apply.
The energy method needs only two numbers: the energy at the start and the energy at the end. Work done = change in energy holds whether the force is constant or not.
Pause and Ponder — Page 1197.1 Work Done by a Constant Force
Pause and Ponder
Q1In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?Fig. 6.8, page 97 (Chapter 6) — redrawn sketch: a weightlifter holding the barbell steady above her head.
Answer
No. She does zero work on the barbell, because the barbell has no displacement.
W = F × s
The upward force she applies = weight of the barbell = mg
Displacement of the barbell, s = 0
W = mg × 0 = 0 J
Q2Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c) — positive, negative or zero?Fig. 6.13, page 99 (Chapter 6): forces acting on the stack of coins — (a) due to the stretched rubber band and friction on release of the rubber band, and (b) due to friction alone, once the coins have left the band and are sliding on the rough surface.
Answer
The work done by friction on the stack of coins is negative.
Direction of motion of the coins →
Direction of the force of friction ← (always opposes relative motion)
Force and displacement are opposite
W = F × (– s) = negative
In-text Questions — Page 1207.2 The Work-Energy Theorem
Q1We have seen that when a force is applied and an object is displaced, work is done on it. Does this cause the object to gain capacity to do further work?
Answer
Yes. Positive work done on an object gives it energy, and energy is the capacity to do work.
A fielder does work on a cricket ball while throwing it. The moving ball then knocks the wicket over — it has done work on the wicket (Fig. 7.8a, b).
Someone does work in raising a flowerpot to a window ledge. If the pot falls, it can damage what lies below — again it does work (Fig. 7.8c).
work done on the object = change in its energy (Eq. 7.3)
positive work → energy gained → capacity to do work on something else
Pause and Ponder — Page 1217.2 The Work-Energy Theorem
Pause and Ponder
Q3When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?
Answer
The chemical energy in your muscles is converted into these forms:
Kinetic energy of the bicycle and of yourself — while you are speeding up. Once you ride at a steady speed this stops growing.
Thermal energy — work done against friction at the axles, the chain and the tyre–road contact, and against air resistance, heats the parts, the road surface and the air.
Sound energy — the small noise of the chain, the tyres and the wind.
Thermal energy inside your body — muscles are not perfectly efficient, so you warm up and sweat.
In-text Questions — Page 1217.4.1 Kinetic energy
Q1How much is the energy possessed by an object by virtue of its motion?
Answer
An object of mass m moving with speed v has kinetic energy K = ½mv2.
Take an object of mass m pushed by a constant force F from speed u to speed v over displacement s.
Kinematics: v2 = u2 + 2as, so s = (v2 – u2) / 2a
Work done: W = F × s = ma × (v2 – u2)/2a
W = ½m(v2 – u2) (Eq. 7.5)
By the work–energy theorem this is the change in the object’s energy.
Starting from rest (u = 0), the whole of it is the kinetic energy gained: K = ½mv2 (Eq. 7.6)
Pause and Ponder — Page 1237.4.1 Kinetic energy
Pause and Ponder
Q4Two objects A and B of mass m and 4 m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B?
Answer
The ratio is vA : vB = 2 : 1.
K = ½mv2
Given KA = KB
½ × m × vA2 = ½ × 4m × vB2
m cancels: vA2 = 4 vB2
Taking the square root: vA = 2 vB vA : vB = 2 : 1
Q5Does the kinetic energy of an object which moves with constant velocity change with its position?
Answer
No. Its kinetic energy stays exactly the same wherever it is.
K = ½mv2
The formula contains only m and v — no h, no x, no position of any kind.
Constant velocity → v does not change → K does not change
Activity 7.1: Let us investigate — Page 1257.4.2 Potential energy — Gravitational Potential Energy
Activity
Q1Raise the ball over the sand bed to a height of about 1 m and drop it (Fig. 7.17). Is a depression created in the sand? Why does the ball create a depression?Fig. 7.17, page 125 — redrawn sketch: depressions created in loose sand by a ball dropped from different heights.
Answer
Yes, a bowl-shaped depression forms. The ball creates it because it arrives with kinetic energy and uses that energy to do work on the sand, pushing the grains aside.
Q2Now, raise the ball to the height of 2 m and release it at a slightly different position over the sand bed such that the depressions do not overlap. Repeat this step one more time. Compare the depths of the depressions. Is there any difference? In which case is the depression deepest and in which case the shallowest?
Answer
Yes, there is a clear difference. The depression made by the ball dropped from 2 m is the deeper one; the one from 1 m is the shallowest. Dropping again from 2 m gives a depression of about the same depth as the first 2 m drop.
Height of drop, h
Potential energy stored, U = mgh
Depth of depression, d = mgh/F
1 m
mg × 1 m
shallowest
2 m
mg × 2 m — twice as much
about twice as deep
Pause and Ponder — Page 1267.4.2 Potential energy
Pause and Ponder
Q6Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?
Answer
Horizontal motion: no change. Vertical raising: the potential energy increases by mgΔh.
U = mgh (a) Horizontal motion: the height h above the ground does not change
ΔU = mg × 0 = 0 J — no change
(b) Raised gradually through Δh:
ΔU = mg × Δh — the potential energy increases
e.g. a 2 kg book lifted 1.5 m: ΔU = 2 kg × 10 m s–2 × 1.5 m = 30 J
Activity 7.2: Let us experiment — Page 1277.4.3 Conservation of mechanical energy
Activity
Q1Take the bob to one side to a point P, which is at the level of the horizontal line and let it go. Observe it at the extreme points of the first couple of oscillations. Does the bob almost reach the level of the horizontal line?
Answer
Yes — on the first few swings the bob rises to almost the horizontal line on the far side, but never quite reaches it, and it falls a little further short with every swing.
Point
Height above lowest point
Potential energy
Kinetic energy
Mechanical energy
P (release)
h
mgh
0
mgh
Q (lowest)
0
0
½mv2 = mgh
mgh
R (far side)
h (almost)
mgh
0
mgh
Pause and Ponder — Page 1297.4.3 Conservation of mechanical energy
Pause and Ponder
Q7For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.Fig. 7.19, page 126: an object of mass m dropped from rest at A (height h), passing B (height h′) after time t, and reaching the ground at C.
Answer
Just before it reaches C the ball has no potential energy left and kinetic energy mgh, so its mechanical energy is still mgh.
The object is released from rest at A, at height h. Take the ground (C) as h = 0.
Height at C: h′ = 0, so potential energy = mg × 0 = 0 J
Speed at C — using v2 = u2 + 2gh with u = 0:
v2 = 0 + 2gh = 2gh
Kinetic energy at C:
K = ½mv2 = ½ × m × 2gh = mgh
Mechanical energy at C = K + U = mgh + 0 = mgh ✔
Q8You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?Fig. 7.22, page 129 — redrawn sketch of the ball roller-coaster exhibit: the ball is released at the top of the tall tower, rises to peak A, dips to B, and then over the successively lower peaks C, D and E.
Answer
The ball is released from rest at the top of the tall tower, so its whole mechanical energy is potential energy there. After that the two forms keep swapping — but the total slowly leaks away.
Point
Height
Potential energy
Kinetic energy
A — first hump
high, but below the start
large
small (only the part of the drop already made)
B — valley between humps
lowest, about ground level
almost zero — minimum
maximum — the ball is fastest here
C — next hump
lower than A
large again, but less than at A
more than at A
Why C, D and E are lower each time — yes, friction is exactly the reason.
Height a ball can climb, h = mechanical energy ÷ mg
On each stretch of track, friction and air resistance do negative work
mechanical energy after = mechanical energy before – (energy lost to friction)
Less mechanical energy → smaller maximum height on the next hump
In-text Questions — Page 1317.6.2 Inclined plane
Q1You can place the box on a smooth inclined plank as shown in (Fig. 7.26b), and push the box up the plank to the platform. Does this method require a smaller force?Fig. 7.26, page 131 — redrawn sketch: a box being (a) lifted vertically up onto the platform, and (b) pushed up a ramp to the same platform.
Answer
Yes. Pushing the box up a smooth ramp needs a smaller force than lifting it straight up — smaller by the factor h/L.
Lifting straight up: force needed = weight = mg, distance = h
Pushing up a ramp of length L to the same height h at constant speed:
work you do = F′ × L
potential energy gained = mgh
Work–energy theorem (ignoring friction): F′ × L = mgh F′ = mgh / L = mg × (h/L)
Since L > h, the factor h/L is less than 1, so F′ < mg
Activity 7.3: Let us experiment — Page 1317.6.2 Inclined plane
Activity
Q1Next, place the plank against the top of the pile of books or stool as shown in Fig. 7.27a. Pull the cart along the plank slowly and steadily. Is the reading of the spring balance (the force required) smaller than that of step 2?Fig. 7.27, page 131 — redrawn sketch: measuring the force needed to pull a cart up an inclined plank of (a) smaller length (steeper) and (b) larger length (less steep), the top of the pile of books being at the same height in both.
Answer
Yes — the spring balance reads clearly less on the ramp than when the cart was lifted straight up.
How the cart is raised
Spring balance reads
Distance moved
Work done
Step 2 — lifted vertically through h = 0.5 m
mg (full weight)
0.5 m
mg × 0.5 m
Step 3 — pulled up a plank of length L = 1.5 m
mg × (h/L) = mg/3, i.e. about one-third
1.5 m
(mg/3) × 1.5 m = mg × 0.5 m — same
Mechanical advantage of the plank = L / h = 1.5 m / 0.5 m = 3
So the effort is one-third of the weight.
If the cart weighs 6 N, the balance should read about 2 N on the ramp.
Q2Now, reduce the angle between the plank and the base as shown in Fig. 7.27b, and repeat the step 3. Observe how the force required changes as the plank becomes less steep.Fig. 7.27, page 131 — redrawn sketch: measuring the force needed to pull a cart up an inclined plank of (a) smaller length (steeper) and (b) larger length (less steep), the top of the pile of books being at the same height in both.
Answer
The force required becomes still smaller as the plank is made less steep — but you must now pull the cart over a longer distance.
For the same height h, a shallower plank means a longer L
Mechanical advantage = L / h → larger
Effort F′ = mg × (h / L) → smaller
Work done = F′ × L = mgh → unchanged
Pause and Ponder — Page 1327.6.2 Inclined plane
Pause and Ponder
Q9Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?Fig. 4.26, page 67 (Chapter 4) — redrawn sketch: a mountain road that winds up the hillside in gentle slopes and hairpin bends.
Answer
A winding hill road is a long inclined plane. It raises a vehicle through the same height h using a much longer path L, so the force the engine must supply along the road is far smaller.
For a slope, mechanical advantage = L / h (Eq. 7.13)
Force needed along the road, F′ = mg × (h / L)
Winding the road makes L large for the same h → F′ becomes small
Work done against gravity = F′ × L = mgh — the same either way
Q10To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.Fig. 7.30, page 132 — redrawn sketch: an inclined ladder and a vertical ladder set against the same building to reach the same height.
Answer
The sloping ladder is an inclined plane. Each step lifts you through a smaller height, so each push needs a smaller force — and the ladder itself supports part of your weight.
Vertical ladder: every step raises you the full rung spacing
→ each push must supply almost the whole weight mg, using your arms as well as your legs
Inclined ladder of length L reaching height h:
force along the ladder, F′ = mg × (h / L) < mg
Total work done = F′ × L = mgh — the same for both ladders
Activity 7.4: Let us investigate — Page 1337.6.3 Lever
Activity
Q1On the other end of the scale, place one eraser. Does the stapler lift up? If not, add one more eraser.
Answer
With one or two erasers the far end of the scale goes down and the heavy stapler is lifted — even though a stapler weighs several times as much as an eraser. The scale works as a lever with the pencil as the fulcrum: the erasers sit far from the fulcrum (long effort arm) and the stapler sits close to it (short load arm), so a small effort balances a large load (effort × effort arm = load × load arm).
In-text Questions — Page 1337.6.3 Lever
Q1By applying a small force at one end of the lever, a larger force can be applied on the object on the other end of the lever. How is this possible?
Answer
It is possible because the lever passes on the work, not the force. The end you press moves through a large distance with a small force; the load end moves a small distance with a large force.
A small effort moving through a long arm delivers a large force through a short arm: F₁ × d₁ = F₂ × d₂.
Work put in at the effort end = work delivered at the load end F1 × d1 = F2 × d2 (Eq. 7.14)
Rearranging: F2 = F1 × (d1 / d2)
Mechanical advantage = effort arm / load arm = d1 / d2 (Eq. 7.16)
Activity 7.5: Let us experiment — Page 1347.6.3 Lever
Activity
Q1Record all observations and measurements, and complete the Table 7.1 by adding more rows. [Number of coins in the left pan, n₁ (Effort); Distance of left pan from the fulcrum, L₁ (cm); Number of coins in right pan, n₂ (Load); Distance of right pan from the fulcrum, L₂ (cm)]
Table 7.1: Number of coins in the left pan and its distance from the fulcrum
Number of coins in left pan, n1 (Effort)
Distance of left pan from the fulcrum, L1 (cm)
Number of coins in right pan, n2 (Load)
Distance of right pan from the fulcrum, L2 (cm)
1
1
1
2
Table 7.1, page 134 — the two rows printed in the book. Copy it out and add more rows for 4 coins and 8 coins in the right pan.
Answer
Keep the left pan (1 coin) at a fixed distance — say L1 = 20 cm — and slide the loaded right pan until the beam is level each time. A typical completed table:
Coins in left pan, n₁ (effort)
Distance L₁ (cm)
Coins in right pan, n₂ (load)
Distance L₂ (cm)
n₁ × L₁
n₂ × L₂
1
20
1
20
20
20
1
20
2
10
20
20
1
20
4
5
20
20
1
20
8
2.5
20
20
The last two columns are always equal, so the beam balances when n1 × L1 = n2 × L2
that is, effort × effort arm = load × load arm (Eq. 7.15)
Mechanical advantage = load / effort = n2/n1 = L1/L2
Row 4: MA = 8/1 = 20 cm / 2.5 cm = 8
Pause and Ponder — Page 1357.6.3 Lever — Classes of levers
Pause and Ponder
Q11Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?Fig. 7.35, page 135 — redrawn sketch: the tip of a spoon is pushed under the lid of a can and the handle is pressed, while the other hand holds the can.
Answer
The spoon works as a class-I lever with a very large mechanical advantage: the rim of the can is the fulcrum, the edge of the lid is the load and your hand at the far end of the handle supplies the effort.
Load arm = distance from the can’s rim to the lid edge ≈ 1 cm
Effort arm = distance from the rim to your hand ≈ 10 cm
Mechanical advantage = effort arm / load arm = 10 cm / 1 cm = 10
So a hand force of 20 N produces a lifting force of
F2 = F1 × (d1/d2) = 20 N × 10 = 200 N on the lid
Q12Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?
Answer
Because moving the object towards the fulcrum shortens the load arm, and a shorter load arm means a larger mechanical advantage — so the blades press on the object with a much bigger force.
Q13Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.
Answer
Every real machine has friction and air resistance acting on its moving parts. These forces do negative work continuously, draining the machine’s mechanical energy into heat and sound — forms that cannot climb back into the moving parts. With no fuel or electricity to top up the store, the mechanical energy runs down to zero and the machine stops.
Work–energy theorem applied to the machine:
change in mechanical energy = work done by all forces
Friction and drag always oppose the motion, so their work is negative
mechanical energy after one cycle = mechanical energy before – (energy lost as heat and sound)
Each cycle the store is smaller → motion becomes slower → finally zero
What If … — Page 1357.6.3 Lever — Classes of levers
What If …
Q1… it were possible to build a perpetual motion machine, which once started, could continue doing useful work forever, without any fuel or electricity?
Answer
It cannot be built, and the reason is the very theorem this chapter is built on: work done = change in energy. Doing useful work forever would mean handing out energy forever from a store that is never refilled.
Useful work delivered in time t = W
Energy taken from the machine’s store = W (nothing else supplies it)
After time t: energy left = E0 – W
As t grows, the store empties and the machine must stop.
Running forever would need E0 = infinite, or energy created from nothing.
Q1State whether True or False. (i) Work is said to be done when a force is applied, even if the object does not move. (ii) Lifting a bucket vertically upward results in positive work done on the bucket. (iii) The SI unit for both work and energy is joule (J). (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.
Answer
#
Statement
True / False
Reason
(i)
Work is done when a force is applied, even if the object does not move
False
W = F × s. With s = 0 the work is zero, however hard you push (Fig. 7.5, pushing a wall).
(ii)
Lifting a bucket vertically upward does positive work on the bucket
True
Your force is upward and the displacement is upward — same direction, so W is positive.
(iii)
The SI unit for both work and energy is the joule (J)
True
Work done appears as a change in energy, so the two must share a unit. 1 J = 1 N m = 1 kg m² s⁻².
(iv)
A motionless stretched rubber band has kinetic energy
False
K = ½mv² and v = 0, so K = 0. It stores elastic potential energy because of its deformation.
(v)
Energy can change from one form to another
True
Electrical → light in a bulb, chemical → mechanical in muscles, mechanical → sound in a bell (Section 7.3).
Q2Fill in the blanks. (i) Work done = ______ × ______ (in the direction of force). (ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force. (iii) The expression for kinetic energy of a body of mass m and velocity v is ______. (iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ______. (v) Power is defined as the ______ at which work is done.
Answer
#
Answer
Where it comes from
(i)
force × displacement
Eq. 7.1 / 7.2: W = F × s
(ii)
1
1 J = 1 N × 1 m, page 118
(iii)
½mv²
Eq. 7.6
(iv)
mgh
Eq. 7.8
(v)
rate
Eq. 7.11: P = W/t
Q3When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct? (i) The force acting on the ball is zero. (ii) The acceleration of the ball is zero. (iii) Its kinetic energy is zero. (iv) Its potential energy is maximum.
Answer
The correct statements are (iii) and (iv).
#
Statement
Correct?
Reason
(i)
The force acting on the ball is zero
No
Gravity never switches off. The weight mg still acts downward at the top.
(ii)
The acceleration of the ball is zero
No
a = F/m = mg/m = g = 10 m s⁻² downward, even at the instant v = 0.
(iii)
Its kinetic energy is zero
Yes
At the highest point v = 0, so K = ½mv² = 0.
(iv)
Its potential energy is maximum
Yes
h is greatest there, so U = mgh is greatest.
Q4For each of the following situations, identify the energy transformation that takes place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, and (ix) a solar panel.
Answer
#
Situation
Energy transformation
What actually happens
(i)
A truck moving uphill
Chemical → kinetic + gravitational potential (+ thermal)
Diesel burns; the engine drives the wheels and the truck also gains height, so it gains mgh. Friction and the hot exhaust carry away the rest.
(ii)
Unwinding of a watch spring
Elastic potential → mechanical (kinetic)
The wound spring is deformed and stores energy; as it unwinds it turns the gears and the hands.
(iii)
Photosynthesis in green leaves
Light (solar) → chemical
Sunlight absorbed by chlorophyll is stored in the bonds of glucose — the food chain’s energy store.
(iv)
Water flowing from a dam
Gravitational potential → kinetic (→ electrical in the turbine)
Water high in the reservoir has mgh; falling turns it into ½mv², which spins the turbine.
(v)
Burning of a matchstick
Chemical → thermal + light
The chemicals on the head react and release stored bond energy as heat and a flame.
(vi)
Explosion of a fire cracker
Chemical → thermal + light + sound + kinetic
A very fast reaction; the hot gases push outward, so fragments and air also gain kinetic energy.
(vii)
Speaking into a microphone
Sound → electrical
Air vibrations move a diaphragm, and the moving diaphragm generates a matching electrical signal.
(viii)
A glowing electric bulb
Electrical → light + thermal
Current heats the filament until it glows. In a filament bulb most of the energy leaves as heat, not light.
(ix)
A solar panel
Light (solar) → electrical
Photocells convert sunlight directly into an electric current.
Q5A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s–2, and student’s mass is m = 50 kg. (i) Find the gain in the potential energy if the student is lifted straight up to the top. (ii) Find the gain in the potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about the dependence of the potential energy on the path taken?
Answer
(i) 36,250 J. (ii) 36,250 J — exactly the same. (iii) Gravitational potential energy depends only on the height gained, not on the path.
(i) By elevator
Formula: ΔU = mgh
ΔU = 50 kg × 10 m s–2 × 72.5 m
ΔU = 36,250 J = 3.625 × 104 J (about 36.25 kJ)
(ii) By the staircase
The starting height and the finishing height are the same, so h is still 72.5 m
ΔU = 50 kg × 10 m s–2 × 72.5 m = 36,250 J — no difference
Q6A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
Answer
Twice the energy is needed — that is 100 % more. The power needed is exactly the same — no extra power at all.
Let the height of one floor be h0 and the first lift take time t.
First lift — to the 10th floor
height = 10h0
E1 = mg × 10h0
P1 = E1/t = 10mgh0 / t
Second lift — to the 20th floor, in time 2t
height = 20h0
E2 = mg × 20h0 = 2E1
P2 = E2/(2t) = 20mgh0 / 2t = 10mgh0 / t = P1
Extra energy required = E2 – E1 = 10mgh0, i.e. 100 % more
Extra power required = P2 – P1 = 0 W
Q7Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
Answer
The energy depends on the mass of the flag, the height of the pole and g. Speed does not change the work done. Doubling the speed doubles the power required.
Factors that fix the energy
E = mgh, so it depends on:
• m — the mass of the flag (and of the rope raised with it)
• h — the height of the flagpole
• g — the acceleration due to gravity
A fixed pulley has mechanical advantage 1; it only turns your downward pull into an upward lift, so it changes neither the force nor the energy.
Slowly or quickly?
W = mgh contains no t. The work done is the same either way.
Doubling the speed
P = W/t. If the speed doubles, the time halves: t′ = t/2
P′ = W / (t/2) = 2W/t = 2P — the power requirement doubles
Q8A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
Answer
The ratio of fuel used on day 1 to day 2 is 4 : 5.
Day 1 — man + scooter
total mass M1 = 60 kg + 100 kg = 160 kg
starting from rest, final speed v:
E1 = ½M1v2 = ½ × 160 kg × v2 = 80v2 joule (v in m s–1)
Day 2 — man + son + scooter
total mass M2 = 60 kg + 40 kg + 100 kg = 200 kg
E2 = ½M2v2 = ½ × 200 kg × v2 = 100v2 joule
Ratio of fuel used (all the energy comes from fuel, with no other losses)
E1 : E2 = 80v2 : 100v2 = 80 : 100 = 4 : 5
Q9On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
Answer
For balance, the child must sit twice as far from the fulcrum as the adult. If the adult sits at a distance d, the child sits at 2d.
A seesaw is a class-I lever. The lighter child sits twice as far from the fulcrum as the adult who weighs twice as much.
Balance condition for a lever (Eq. 7.15):
effort × effort arm = load × load arm
Let the child’s weight be W and the adult’s be 2W.
W × dchild = 2W × dadult
W cancels: dchild = 2 × dadult
So if the adult sits 1 m from the fulcrum, the child must sit 2 m from it.
Check: 300 N × 2 m = 600 N m and 600 N × 1 m = 600 N m ✔
Q10A ball of mass 2 kg is thrown up with a velocity of 20 m s–1. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s–2).
Answer
(i) Negative going up, positive coming down. (ii) The air resistance did –12 J of work on the ball during the rise.
(i) Gravity acts downward. Going up, the displacement is opposite to it → negative work. Coming down, the displacement is along it → positive work.
(ii) Kinetic energy at the throw, Ki = ½ × 2 kg × (20 m s–1)2 = 400 J; at the top, Kf = 0 J
Work done by gravity = –mgh = –(2 kg × 10 m s–2 × 19.4 m) = –388 J
Work–energy theorem: Wgravity + Wair = Kf – Ki
–388 J + Wair = 0 J – 400 J
Wair = –12 J
Q11A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block’s speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?Fig. 7.37, page 138: the force applied on the block against its displacement.
Answer
(i) 6 m s⁻¹ at 0 m. (ii) About 8.12 m s⁻¹ at 4 m. The acceleration is never negative.
(i) K = ½mv2 → v = √(2K/m) = √(2 × 180 J / 10.0 kg) = √36 m s–1 = 6 m s–1
Work done by the force from 0 m to 4 m = area under the graph (a trapezium)
W = ½ × (4 m + 2 m) × 50 N = 150 J
(ii) Work–energy theorem: K at 4 m = 180 J + 150 J = 330 J
v = √(2 × 330 J / 10.0 kg) = √66 m s–1 ≈ 8.12 m s–1
Negative acceleration? No. The force is positive (along the motion) all the way from 0 m to 4 m, so a = F/m is never negative. Between 3 m and 4 m it only decreases, from 5 m s–2 to 0.
Q12The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
Answer
The ball will rise to 48 m on the Moon — six times as high.
The astronaut throws with the same speed u both times, so the ball starts with the same kinetic energy.
At the highest point all of it has become potential energy:
½mu2 = mgh → h = u2 / 2g
On the Earth
hE = u2 / (2gE) = 8 m
On the Moon, gM = gE/6
hM = u2 / (2gM) = u2 / (2 × gE/6) = 6 × u2/(2gE) = 6 hE
hM = 6 × 8 m = 48 m
Q13A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?Fig. 7.38, page 138: speed–time graph of the car from the instant the driver spots the obstruction (A) to the halt (C).
Answer
(i) From A to B (0 s to 1 s) the graph is flat: the car moves at a constant speed of 35 m s–1, with zero acceleration. The brakes have not yet been applied — this one second is the driver’s reaction time.
(ii) K at A = ½mv2 = ½ × 1000 kg × (35 m s–1)2 = 6,12,500 J
(iii) At B, K = 6,12,500 J (the speed is still 35 m s–1); at C, K = 0 J
Work done by the brakes = change in kinetic energy = 0 J – 6,12,500 J = –6,12,500 J
(negative, because the braking force opposes the motion)
(iv) The kinetic energy is transformed mainly into thermal energy — heat in the brake pads and discs, in the tyres and in the road surface — together with a small amount of sound energy (the screech of the tyres).
Q14The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s–1 and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.Fig. 7.39, page 138: potential energy of the 0.5 kg ball against its displacement along the frictionless track.
Answer
At P the ball moves at about 6.3 m s⁻¹, at Q it is momentarily at rest (0 m s⁻¹), and it can never reach R at all.
Step 1 — find the total mechanical energy
At O: v = 0 m s–1, so K = 0 J; and U = 30 J
Total mechanical energy E = K + U = 0 + 30 = 30 J
The track is frictionless, so E stays 30 J everywhere.
Step 2 — at each point, K = E – U, then v = √(2K/m), with m = 0.5 kg
Point
U from the graph
K = 30 J – U
v = √(2K/m)
P
20 J
10 J
√(2 × 10 / 0.5) = √40 ≈ 6.3 m s⁻¹
Q
30 J
0 J
0 m s⁻¹ — momentarily at rest
R
40 J
–10 J
impossible — the ball never gets to R
Q15A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s–2.
Answer
(i) About 14.1 m s⁻¹. (ii) The depression is about 0.05 m, that is 5 cm deep.
(i) Mechanical energy is conserved during the fall: mgh = ½mv2
v2 = 2gh = 2 × 10 m s–2 × 10 m = 200 m2 s–2
v = √200 m s–1 ≈ 14.1 m s–1
(ii) Energy of the coconut on reaching the sand = mgh = 1.5 kg × 10 m s–2 × 10 m = 150 J
The sand removes all of it: F × d = 150 J
d = 150 J ÷ 3000 N = 0.05 m = 5 cm
The Journey Beyond — Page 139Project work
The Journey Beyond
Q1Remove both the ends from a pen so that the refill can slide freely through the barrel (Fig. 7.40). Fix the pen cap to the side of the barrel and attach a rubber band to the clip of the cap. Connect the free end of the rubber band to the refill using a safety pin. Stretch and release the rubber band. The refill shoots out, showing the conversion of elastic potential energy into kinetic energy. Repeat with different amounts of stretch, and observe how the distance travelled changes. Is there a relationship between the stretch and the distance travelled?Fig. 7.40, page 139 — redrawn sketch of the pen launcher: the cap is fixed to the side of the barrel and a rubber band runs from the cap’s clip to a safety pin attached to the refill, which slides freely inside the barrel.
Answer
Yes. The more the rubber band is stretched, the farther the refill travels. A larger stretch stores more elastic potential energy in the band. On release this energy becomes the kinetic energy of the refill, so the refill leaves faster and goes farther.
Stretch of the band (cm)
Distance travelled by the refill (cm) — sample readings
1
25
2
60
3
110
Conclusion: the distance travelled increases as the stretch increases. Your own readings will differ, but the trend will be the same.
Q2Construct one or more simple machines, or a combination of them (lever, pulley, and inclined plane) using easily available materials, such as cardboard, wooden strips or rulers, pencils or bolts (to act as a fulcrum), thread or rope, small pulleys (or two bottle caps stuck together), and paper cups to hold small weights. Be imaginative in your design. Use your model to lift or move a small load, measure the effort and the load, and calculate the mechanical advantage.
Answer
Build any one of these three, then measure the effort with a spring balance (or by counting identical coins) and use MA = load ÷ effort.
Machine
How to build it
What to measure
Expected mechanical advantage
Lever
A 30 cm ruler resting across a pencil. Load (a stone in a paper cup) near the pencil, effort at the far end.
Load arm d₂, effort arm d₁, and the effort needed
MA = d₁/d₂; with d₁ = 24 cm and d₂ = 6 cm, MA = 4
Inclined plane
A stiff cardboard ramp of length L from the floor to a stack of books of height h. Pull a toy car up with a spring balance.
L, h, and the spring balance reading
MA = L/h; with L = 60 cm and h = 15 cm, MA = 4
Pulley
Two bottle caps glued back to back on a nail axle, hung from a hook. Thread over the groove, load on one side, effort on the other.
Weight of load, force needed on the free end
MA ≈ 1 for a single fixed pulley — it changes only the direction
Sample answer — the ruler lever
Load = a 200 g stone, so load = mg = 0.2 kg × 10 m s–2 = 2 N
Load arm d2 = 6 cm, effort arm d1 = 24 cm
Predicted effort: F1 = F2 × d2/d1 = 2 N × 6/24 = 0.5 N
Measured effort (spring balance) = 0.6 N
Mechanical advantage = load ÷ effort = 2 N ÷ 0.6 N = 3.3
Ideal MA = d1/d2 = 24/6 = 4
Q3Computer simulations can help in visualising physical quantities that are difficult to observe directly. The PhET simulations (https://phet.colorado.edu) provide interactive models, such as Energy Skate Park, Energy Forms and Changes, Pendulum Lab, and Masses and Springs. Use these to explore how different forms of energy change as parameters, such as mass, height, and friction are varied.
Answer
Each simulation shows an energy bar chart alongside the motion. Change one quantity at a time and watch which bar grows and which shrinks — that is the whole of this chapter, on screen.
Simulation
What to change
What to look for
Which idea it proves
Energy Skate Park
Set friction to zero; drop the skater from different heights
The KE and PE bars swap size, but the total bar never moves
Conservation of mechanical energy (Section 7.4.3)
Energy Skate Park
Now turn friction on
A thermal-energy bar appears and grows; the skater’s highest point falls each pass
Why the roller-coaster humps get lower (Question 8, page 129)
Energy Skate Park
Change the skater’s mass, keeping the height
All bars scale up together, but the speed at the bottom is unchanged
v = √(2gh) does not depend on mass (Think It Over, page 116)
Pendulum Lab
Release the bob from different angles, with and without friction
Without friction it returns to the same height every swing; with friction, a little less each time
Activity 7.2, page 127
Masses and Springs
Stretch the spring by different amounts
The elastic-PE bar grows much faster than the stretch
Energy stored by deformation (Section 7.4.2)
Energy Forms and Changes
Connect the water wheel, the generator and the bulb
Energy symbols pass along the chain; some always leak away as heat
Forms of energy, Fig. 7.10; the gharat on page 136