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 ___.

Answer 1

The work done on the ball during each revolution can be calculated using the formula for work. In this case, the force acting on the ball is the tension in the string, and the distance is the circumference of the circle. Plugging in the given values, the work done on the ball is approximately 226.35 J.

Step-by-step explanation:

The work done on the ball during each revolution can be calculated using the formula for work:

Work = Force x Distance

In this case, the force acting on the ball is the tension in the string, and the distance is the circumference of the circle. The tension in the string can be calculated using the centripetal force equation:

Tension = (mass x velocity^2) / radius

Using the given values:Mass = 2 kg ,Velocity = 6 m/s, Radius = 0.80 m

Plugging these values into the equation, we get: Tension = (2 kg x (6 m/s)^2) / 0.80 m = 45 N

The distance traveled by the ball during each revolution is equal to the circumference of the circle:Distance = 2 x π x radius

Plugging in the given value for the radius:Distance = 2 x π x 0.80 m ≈ 5.03 m

Now we can calculate the work done on the ball using the formula:Work = Tension x Distance

Plugging in the values we calculated:Work = 45 N x 5.03 m ≈ 226.35 J

Therefore, the work done on the ball during each revolution is approximately 226.35 J.