A 4.40-kg block is set into motion up an inclined plane with an initial speed of

Question

A 4.40-kg block is set into motion up an inclined plane with an initial speed of v_i = 8.60 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 theta = 30.0degree to the horizontal. (a) For this motion, determine the change in the block's kinetic energy. (b) For this motion, determine the change in potential energy of the block-Earth system. (c) Determine the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?

Explanation / Answer

The total KE at start =0.5m*V0^2=0.5*4.4*(8.6)^2 =162.71 N-m.
(a) As the Final KE=0, Net change in KE is 162.71 N-m
(b) PE at start is zero. PE at destination = m*g*h= 4.4*9.8*3*sin30= 64.68 N-m.

So this is the total Change in PE.
(c) Ideally, whole of KE should have converted to PE.
So energy lost to friction =KE-PE=162.71 – 64.68 = 98.03 N-m.
As energy = F.d, = F*3 = 98.03
So Force of friction F= 98.03/3= 32.68 N.
(d) For 4.4 Kg at 30 deg, Normal Force N = 4.4 *Cos30*9.8 = 37.34
So Coeff. of Kinetic Friction = 32.68/37.34 = 0.875

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