NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy And Power

Ans : (a) Positive 
In the given case, force and displacement are in the same direction. Hence, the sign of 
work done is positive. In this case, the work is done on the bucket. 
(b) Negative 
In the given case, the direction of force (vertically downward) and displacement 
(vertically upward) are opposite to each other. Hence, the sign of work done is negative. 
(c) Negative 
Since the direction of frictional force is opposite to the direction of motion, the work done 
by frictional force is negative in this case. 
(d) Positive 
Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive. 
(e) Negative 
The resistive force of air acts in the direction opposite to the direction of motion of the 
pendulum. Hence, the work done is negative in this case. 

Ans : \(\begin{array}{l}{\text { Mass of the body, } m=2 \mathrm{kg}} \\ {\text { Applied force, } F=7 \mathrm{N}} \\ {\text { Coefficient of kinetic friction, } \mu=0.1} \\ {\text { Initial velocity, } u=0} \\ {\text { Time, } t=10 \mathrm{s}}\end{array}\)
\(\begin{array}{l}{\text { The acceleration produced in the body by the applied force is given by Newton's seconc }} \\ {\text { law of motion as: }} \\ {a^{\prime}=\frac{F}{m}=\frac{7}{2}=3.5 \mathrm{m} / \mathrm{s}^{2}}\end{array}\)
\(\begin{array}{l}{\text { Frictional force is given as: }} \\ {f=\mu m g} \\ {=0.1 \times 2 \times 9.8=-1.96 \mathrm{N}} \\ {\text { The acceleration produced by the frictional force: }}\end{array}\)
\(\begin{array}{l}{a=a^{\prime}+a^{\prime \prime}} \\ {\quad=3.5+(-0.98)=2.52 \mathrm{m} / \mathrm{s}^{2}} \\ {\text { The distance travelled by the body is given by the equation of motion: }} \\ {s=u t+\frac{1}{2} a t^{2}}\end{array}\)
\(=0+\frac{1}{2} \times 2.52 \times(10)^{2}=126 \mathrm{m}\)
\(\begin{array}{l}{\text { (a) Work done by the applied force, } W_{\mathrm{a}}=F \times \mathrm{s}=7 \times 126=882 \mathrm{J}} \\ {\text { (b) Work done by the frictional force, } W_{f=} F \times s=-1.96 \times 126=-247 \mathrm{J}} \\ {\text { (c) Net force }=7+(-1.96)=5.04 \mathrm{N}}\end{array}\)
\(\begin{array}{l}{\text { Work done by the net force, } W_{\text { net }}=5.04 \times 126=635 \text { J }} \\ {\text { (d) From the first equation of motion, final velocity can be calculated as: }} \\ {v=u+a t} \\ {=0+2.52 \times 10=25.2 \mathrm{m} / \mathrm{s}}\end{array}\)
\(\begin{array}{l}{\text { Change in kinetic energy }=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}} \\ {=\frac{1}{2} \times 2\left(v^{2}-u^{2}\right)=(25.2)^{2}-0^{2}=635 \mathrm{J}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { (a) } x > a ; 0} \\ {\text { Total energy of a system is given by the relation: }}\end{array}\)
\(\begin{array}{l}{E=P . E .+\mathrm{K} . \mathrm{E}} \\ {\therefore \mathrm{K} \cdot \mathrm{E} .=E-\mathrm{P} . \mathrm{E}}\end{array}\)
Kinetic energy of a body is a positive quantity, It cannot be negative. Therefore, the 
particle will not exist in a region where K.E. becomes negative. 
In the given case, the potential energy of the particle becomes greater than total 
energy (E) for x > a. Hence, kinetic energy becomes negative in this region. Therefore, 
the particle will not exist is this region. The minimum total energy of the particle is zero. 
(b) All regions 
In the given case, the potential energy (Vo) is greater than total energy (E) in all 
regions. Hence, the particle will not exist in this region. 
\((\mathrm{c}) x  >a \text { and } x < b ;-V_{1}\)
In the given case, the condition regarding the positivity of K.E. is satisfied only in the 
region between x > a and x < b. 
The minimal potential energy in this case is —VI. Therefore, K E 
Therefore, for the positivity of the kinetic energy, the total energy of the particle must 
be greater than —VI. So, the minimum total energy the particle must have is —VI. 
(d)\(-\frac{b}{2} < x < \frac{a}{2} ; \quad \frac{a}{2} < x < \frac{b}{2} ;-V_{1}\)
In the given case, the potential energy (Vo) of the particle becomes greater than the 
total energy (E) for \(-\frac{b}{2} < x < \frac{b}{2} \text { and }-\frac{a}{2} < x < \frac{a}{2}\)
Therefore, the particle will not exist in these region.
The minimum potential energy in this case is —VI. Therefore, K.E. E — (—VI) E + VI. 
Therefore, for the positivity of the kinetic energy, the total energy of the particle must 
be greater than -VI. So, the minimum total energy the particle must have is -VI. 

Ans : \(\begin{array}{l}{\text { Total energy of the particle, } E=1 \mathrm{J}} \\ {\text { Force constant, } k=0.5 \mathrm{Nm}^{-1}} \\ {\text { Kinetic energy of the particle, } \mathrm{K}=\frac{1}{2} m v^{2}} \\ {\text { According to the conservation law: }}\end{array}\)
\(\begin{array}{l}{E=V+K} \\ {1=\frac{1}{2} k x^{2}+\frac{1}{2} m v^{2}} \\ {\text { At the moment of 'turn back', velocity (and hence } K \text { ) becomes zero. }}\end{array}\)
\(\begin{array}{l}{1=\frac{1}{2} k x^{2}} \\ {\frac{1}{2} \times 0.5 x^{2}=1} \\ {x^{2}=4} \\ {x=\pm 2}\end{array}\)
Hence, the particle turns back when it reaches x ± 2 m 

Ans : (a) Rocket 
The burning of the casing of a rocket in flight (due to friction) results in the reduction of 
the mass of the rocket. 
According to the conservation of energy: 
Total Energy (T .E.) = Potential energy Kinetic energy (ICE.) 
\(=m \mathrm{g} h+\frac{1}{2} m v^{2}\)
The reduction in the rocket's mass causes a drop in the total energy. Therefore, the heat 
energy required for the burning is obtained from the rocket. 
(b) Gravitational force is a conservative force. Since the work done by a conservative 
force over a closed path is zero, the work done by the gravitational force over every 
complete orbit of a comet is zero. 
(c) 'When an artificial satellite, orbiting around earth, moves closer to earth, its potential 
energy decreases because of the reduction in the height. Since the total energy of the 
system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the 
velocity of the satellite increases. However, due to atmospheric friction, the total energy 
of the satellite decreases by a small amount. 
(d) In the second case 
\(\begin{array}{l}{\frac{\text { case (i) }}{\text { Mass, }}} \\ {\text { Mass, } m=15 \mathrm{kg}} \\ {\text { Displacement, } s=2 \mathrm{m}} \\ {\text { Work done, } W=F s \cos \theta} \\ {\text { Where, } \theta=\text { Angle between force and displacement }} \\ {=m g s \cos \theta=15 \times 2 \times 9.8 \cos 90^{\circ}} \\ {=0}\end{array}\)
\(\begin{array}{l}{\text { Case (ii) }} \\ {\text { Mass, } m=15 \mathrm{kg}} \\ {\text { Displacement, } s=2 \mathrm{m}} \\ {\text { Here, the direction of the force applied on the rope and the direction of the displacement }} \\ {\text { of the rope are same. }}\end{array}\)
\(\begin{array}{l}{\text { Therefore, the angle between them, } \theta=0^{\circ}} \\ {\text { since cos } 0^{\circ}=1} \\ {\text { Work done, } W=F s \cos \theta=m g s} \\ {=15 \times 9.8 \times 2=294 \mathrm{J}} \\ {\text { Hence, more work is done in the second case. }}\end{array}\)