Answered

a 14.5 kg box is moving across level ground against friction. the coefficient of kinetic friction is equal to 0.165. the box travels 10.2 m before coming to a complete stop. what was the initial velocity of the box?



Answer :

MathPhys

Answer:

5.74 m/s

Explanation:

The work-energy theorem says that the work done an object is equal to its change in kinetic energy. Work equals force times distance, and for a box on level ground, the friction force equals the weight times the coefficient of friction.

W = ΔKE

Fd = KE

mgμ d = ½ mv²

gμ d = ½ v²

(9.8) (0.165) (10.2) = ½ v²

v = 5.74 m/s

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