A box of mass m is initially at rest at the peak of an inclined plane, which has

Question

A box of mass m is initially at rest at the peak of an inclined plane, which has a height of 4.8 m and has an angle of 21 degrees with respect to the horizontal. After letting go of the object, it is found to be traveling at v= 0.85 m/s a distance d after the end of the inclined plane. The coefficient of kinetic friction between the box and the plane is 0.1, and the coefficient of friction on the horizontal surface is 0.2. What is the speed of the box, in meters per second, just after it leaves the inclined plane? Find the distance, d, in meters.

Can someone show me how to set up this problem

Explanation / Answer

the potential energy = mgh

at the end of inclined plane velocity = 0,85 m/s

now

the distance on the plane = 4.8/ sin( 21) = 13.39 m

using the conservation of energy

potential energy = work done by friction + kinetic energy at bottom

mgH= mu* mg * cos(21) * 13.39 + 0.5 m*v2......................(1)

now, the kinetic energy at the bottom of inclined plane = kinetic energy at distance d + work doen by froction

so,

0.5 m*v2 = 0.5 m*(0.85)2 + mu' * mg .....................(2)

put (2) in (1)

we get,

mgH= mu* mg * cos(21) * 13.39 + 0.5 m*(0.85)2 + mu' * mg*d

=> 9.8* 4.8= 0.1*9.8 *0.933 *13.39 + 0.36 + 0.1* 9.8* d

=> d = 35.139 m

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