Question

• An object of mass 2.0 kg moves in uniform circular motion on a horizontal friction less table. The radius of the circle is 0.75 m and the centripetal force is 10.0 N. The work done by this force when the object moves through one-half of a complete revolution is​

Answer 1

Answer:

23.562

Explanation:

First it is good to work out the circumference:

2 pi r

2 pi .75

circumference = 4.7124

Next we need half of this because it is half a resolution. This will be our distance travelled.

half = 2.3562

word done = force x distance

work done = 10 x 2.3562

work done = 23.562

(I didn't see the need for mass...)