Question

A 2 kg ball is attached to a 0.80 m string and whirled in a horizontal circle at a constant speed of 6 m/s. The work done on the ball during each revolution is: * 1 point (A) 90 J (B) 72 J (D) 16 J (D) zero

Answer 1

The work done on the ball during each revolution is 0 J. (Option D).

How to calculate the work done?

The work done on the ball during each revolution is calculated by applying the following formula as shown below.

W = Fd

where;

• F is the applied force of the ball
• d is the displacement of the ball

The given parameters include;

The mass of the ball = 2 kg

Radius of the circle = 0.8 m

Speed, v = 6 m/s

The applied force is;

F = mv²/r

F = (2 x 6²) / 0.8

F = 90 N

The displacement of the ball;

1 complete revolution = 0 m displacement

Work done = 90 N x 0 m

Work done = 0 J

Answer 2

The answer should be 36 Joules.

Step-by-step explanation:

This can be solved as a angular motion.

Angular velocity w = V/r

Where V is the linear velocity 6 m/s

And r is the radius of rotation which is equal to the lenght of the string 0.8 m

w = V/r = 6/0.8 = 7.5 rad/s

Moment of inertia of rotation I = mr^2

Where m is the mass of the ball 2 kg

I = 2 x 0.8^2 = 1.28 kg-m2

From energy conservation, work done is equal to the kinetic energy of the angular motion.

KE = 0.5Iw^2

= 0.5 x 1.28 x 7.5^2 = 36 Joules.