A block is released from rest at the top of an inclined 6.20 m long. The angle o

Question

A block is released from rest at the top of an inclined 6.20 m long. The angle of the incline with respect to the horizontal direction is and the coefficient of kinetic friction between the block and the surfaces (incline and horizontal) is . The block slides along the incline with constant velocity and continues moving along the horizontal surface until it comes to rest. Using the work-energy theorem, Determine:

a) The speed reached by the block at the bottom of the incline.   

b) The displacement of the block along the horizontal surface

Explanation / Answer

if block was released from rest from the top position then it will acquire velocity as it comes down which can be calculated using work-energy theorem -

mgh = 0.5 mv2

v2 = 2gh

v2 = 2g*L cos theta

v2 = 2*9.8*6.20 cos theta

v2 = 121.52 cos theta

b> work done against friction willl be equal to kinetic energy of the block at the bottom-

u*mg*s = 0.5*m*121.52 cos theta

s = 6.2 cos theta/u

where u = coefficient of friction

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