A 5.00-kg block is set into motion up an inclined plane with an initial speed of

Question

A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s (sec figure below). The block comes rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0 degree to the horizontal. For this motion determine (a) the change in the block's kinetic energy, (b) the change in the potential energy of the block-Earth system, and (c) the friction force exerted on the block (assumed to be constant). (d) Find the coefficient of kinetic friction using conservation of total energy?

Explanation / Answer

#a) dK = Kf - Ki = 0-0.5*m*v1^2 = -0.5*5*8*8 = -160 J

#b) dPE = U2 - U1 = m*g*L*sin30 - 0 = 5*9.8*3*0.5 = 73.5 J

#c)

from energy conservation

dU + Wf + dK = 0

73.5 + f*3 - 180 = 0

f = 35.5 N

#d)

f = u*N = u*m*g*cos30

u = f/(m*g*cos30)

u = 0.836

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