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

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west