ka A 4.80-kg block is set into motion up an inclined plane with an initial speed
ID: 2115382 • Letter: K
Question
ka
A 4.80-kg block is set into motion up an inclined plane with an initial speed of vi = 8.40 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 = 30.0 degree to the horizontal. For this motion, determine the change in the block's kinetic energy. change in a quantity is defined as the final value minus the initial value. J For this motion, determine the change in potential energy of the block-Earth system. J Determine the friction force exerted on the block (assumed to be constant). N What is the coefficient of kinetic friction?Explanation / Answer
a) change in K.E= 0-.5*4.80*8.4^2=-169.344 J
b)Change in P.E = 169.344
C) Work done = Change in K.e
-(4.8*9.8*sin(30)*3+F*3)= -169.344
F=32.928 N
d)u=32.928/4.8*9.8*cos(30) =.808
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