# Work, Energy and Power

## Class 11 NCERT Physics

### NCERT

1   The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:$\\$ (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.$\\$ (b) work done by gravitational force in the above case,$\\$ (c) work done by friction on a body sliding down an inclined plane,$\\$ (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,$\\$ (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

##### Solution :

(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.

2   A body of mass $2 kg$ initially at rest moves under the action of an applied horizontal force of $7 N$ on a table with coefficient of kinetic friction $= 0.1.$$\\ Compute the\\ (a) work done by the applied force in 10 s,$$\\$ (b) work done by friction in $10 s,$$\\ (c) work done by the net force on the body in 10 s,$$\\$ (d) change in kinetic energy of the body in $10 s,$$\\ and interpret your results. ##### Solution : Mass of the body, m = 2 kg$$\\$ Applied force, $F = 7 N$$\\ Coefficient of kinetic friction, \mu = 0.1$$\\$ Initial velocity,$u = 0$$\\ Time, t = 10 s$$\\$ The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:$\\$ $a ' =\dfrac{F}{m}= \dfrac{7 }{ 2} = 3.5 ms^{-2}$ Frictional force is given as:$\\$ $f = \mu mg\\ = 0.1 × 2 × 9.8 = - 1.96 N$$\\ The acceleration produced by the frictional force:\\ a "=\dfrac{- 1.96 }{ 2}=-0.98 ms^2$$\\$ Total acceleration of the body: $a ' + a "\\ = 3.5+(-0.98)=2.52 ms^{-2}$ The distance travelled by the body is given by the equation of motion: $s=ut +(\dfrac{1}{2}) at^2\\ =0 +(\dfrac{1}{2})* 2.52*(10)^2=126 m$$\\ (a) Work done by the applied force, W_a =F*s =7 *126 =882 J$$\\$ (b) Work done by the frictional force, $W_f =F *s =-1.96 * 126 =- 247 J$$\\ (c) Net force = 7 + (-1.96) = 5.04 N$$\\$ Work done by the net force, $W_{net}= 5.04*126 = 635 J$$\\ (d) From the first equation of motion, final velocity can be calculated as: v = u + at\\ = 0 + 2.52 × 10 = 25.2 m/s Change in kinetic energy=(\dfrac{1}{2})mv^2-(\dfrac{1}{2})mu^2\\ =(\dfrac{1}{2})*2(v^2-u^2)\\ =(25.2)^2-0^2=635 J 3 Given in figure are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. ##### Solution : Total energy of a system is given by the relation:\\ E = P.E. + K. E.\\ \therefore K.E. = E - P.E.$$\\$ 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.$\\$ (i)For $x> a,P.E. (V_0) > E$$\\ \therefore K.E. becomes negative. Hence, the object cannot exist in the region x > a.$$\\$ (ii)For $x < a$ and $x> b,P.E.(V_0) >E.$$\\ \therefore K.E. becomes negative. Hence the object cannot be present in the region x < a and x > b.$$\\$ (iii)$x > a$ 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 minimum potential energy in this case is $- V_1$ . Therefore, $K.E. = E - (- V_1)=E +V_1 .$ Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than $-V_1$. So, the minimum total energy the particle must have is $- V_1 .$ $\\$ (iv)$-\dfrac{b}{2} < x< \dfrac{a}{2};$$\\ \dfrac{a}{2}< x< \dfrac{b}{2};-V_1$$\\$ In the given case, the potential energy $(V_0)$ of the particle becomes greater than the total energy$\\$ (E) For $-\dfrac{b}{2} < x < \dfrac{b}{2}$ and $-\dfrac{a}{2} < x < \dfrac{a}{2}.$\\$Therefore,$ K . E . = E -(- V_1)= E +V_1 .$\\$ Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than $-V_1$ . So, the minimum total energy the particle must have is $-V_1$.

4   The potential energy function for a particle executing linear simple harmonic motion is given by$V(x)=kx^2/2,$ where $k$ is the force constant of the oscillator. For $k=0.5 m^{-1},$ the graph of $V (x)$ versus $x$ is shown in Fig. $6.12.$ Show that a particle of total energy $1 J$ moving under this potential must ‘turn back’ when it reaches $x =\pm 2 m .$

##### Solution :

(a) False $\\$ In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.$\\$ (b) False$\\$ The external forces on the body may change the total energy of the body.$\\$ (c) False$\\$ The work done in the motion of a body over a closed loop is zero for a conservation force only.$\\$ (d) True$\\$ In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.

8   Answer carefully, with reasons:$\\$ (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?$\\$ (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?$\\$ (c) What are the answers to (a) and (b) for an inelastic collision?$\\$ (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

##### Solution :

(a) No$\\$ $K.E.$ is not conserved during the given elastic collision, $K.E.$ before and after collision is the same. In fact during collision, $K.E.$ of the balls gets converted into potential energy.$\\$ (b) Yes$\\$ In an elastic collision, the total linear momentum of the system always remains conserved.$\\$ (c) No; Yes$\\$ In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.$\\$ (d) Elastic$\\$ In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.

9   A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to