Chapter 11, Problem 79A

(a)

To determine

The work done on the mass by the tension of the string in one revolution.

Answer to Problem 79A

  $0.0\text{ J}$

Explanation of Solution

Given:

Mass of an object which is used to swing is $m=55\text{ g = 55}\times {\text{10}}^{-3}\text{ kg}$

Length of the string is $l=0.75\text{ m}$

    Figure 1

Formula used:

Work done on the mass by the tension of the string in one revolution can be calculated using the equation,

  $W=Fl\cos \theta$  $\left(1\right)$

Where, $F$ is force

  $l$ is length of the string

Calculation:

When a mass moves in a circular path, the centripetal force acting on the mass is directed towards the center of the circle and the motion of the mass is at right angle to the tension of the string. Hence $\theta =90°$

Substituting the numerical values in equation $\left(1\right)$ ,

  $W=Fl\cos 90°$

  $=0.0\text{ J}$

Conclusion:

The work done on the mass by the tension of the string in one revolution is $0.0\text{ J}$ .

(b)

To determine

To explain − whether the answer obtained to part (a) is in agreement with the work-energy theorem.

Explanation of Solution

Introduction:

Work-energy theorem: It states that the work done by all the forces exerted on a body is equivalent to the variation in kinetic energy of it. That is,

Work done = Final kinetic energy− initial kinetic energy

Mathematically this can be represented as,

  $W=\Delta KE=K{E}_f-K{E}_i$

Yes. The answer obtained in the part (a) is in agreement with the work - energy theorem. Because the mass rotates with constant kinetic energy. That means initial and final kinetic energy of the rotating mass are equal. And hence the change in kinetic energy of that mass is zero. The mass rotates in a circular path with a continuous speed.