A 5.40–kg block is set into motion up an inclined plane with an initial speed of

Question

A 5.40–kg block is set into motion up an inclined plane with an initial speed of vi = 7.60 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of = 30.0° to the horizontal.

(a) For this motion, determine the change in the block's kinetic energy.
J

(b) For this motion, determine the change in potential energy of the block–Earth system.
J

(c) Determine the friction force exerted on the block (assumed to be constant).
N

(d) What is the coefficient of kinetic friction?

Explanation / Answer

A) change in kinetic energy dK =0 - 0.5*m*vi^2 = -0.5*5.4*7.6*7.6 = -155.952 J

B) chnage in potential energy =dp = m*g*d*sin(30) = 5.4*9.81*3*0.5 = 79.461 J

C) work done by the net force = change in kE+ change in PE = -155.952+79.461 = -76.491 J

-(m*g*sin(30)+fk)*d = -76.491

m*g*sin(30) + fk = 76.491/3 = 25.497

fk = 25.497-(5.4*9.81*0.5) = -0.99 N

D) fk = mu_k*m*g*cos(30)

0.99 = mu_k*5.4*9.81*0.866

mu_k = 0.0215

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