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The kinetic friction force that a horizontal surface exerts on a 60.0-kg object is 50.0 n. if the initial speed of the object is 25.0 m/s, what distance will it slide before coming to a stop?

Respuesta :

Answer:

KE = 1/2 M v^2

KE = 60/2 * 25^2 = 18750 Joules         initial Kinetic Energy

W = Fs * d        distance thru which friction acts

d = W / Fs = 18750 J / 50 N = 375 m

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