A hoop (I = MR2) of mass 2.0 kg and radius 0.50 m is rolling at a center-of-mass speed of 15 m/s. An external force does 750 J of work on the hoop. What is the new speed of the center of mass of the hoop?

24.49 m/s

Explanation:

Given

Mass of hoop, m = 2 kg

Radius of hoop, r = 0.5 m

Speed of rolling, v = 15 m/s

Work done, = 750 J

If we assume that the hoop is not skidding, then it's Kinetic Energy of rotation is 1/2mv²

also the Kinetic Energy of translation is 1/2mv², thus the total Kinetic Energy is

KE = KE(r) + KE(t)

KE = 1/2mv² + 1/2mv²

KE = mv²

Recall, the change in kinetic energy = work done = 750 J

mv² - mv•² = 750

m (v² - v•²) = 750

v² - v•² = 750/m

v² - 15² = 750/2

v² = 375 + 225

v² = 600

v = 24.49 m/s

Therefore, the new speed is 24.49 m/s

the answer is b ) square root of 5