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Lunna [17]
3 years ago
10

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

ork on the hoop. What is the new speed of the center of mass of the hoop (in m/s)? Round your answer to the nearest whole number.
Physics
1 answer:
Scilla [17]3 years ago
3 0

Answer:

v_f = 20 m/s

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2

now for pure rolling condition we will have

v = R\omega

also we have

I = mR^2

now we will have

KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}

KE = mv^2

now by work energy theorem we can say

W = KE_f - KE_i

842 J = mv_f^2 - mv_i^2

842 = 3(v_f^2) - 3\times 11^2

now solve for final speed

v_f = 20 m/s

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