Question

A 150 gram block is pushed up a ramp set at a 20 degree angle with respect to the horizontal. Its initial speed is 1.24 m/s as a result of the push, and it travels 14.0 cm up the incline before coming to rest. How far would it travel down the ramp before coming to rest if the push was in the opposite direction but the initial speed was the same?

Answer #1

Applying Equilibrium condition in the direction normal to the displacement:

N = mg cos(20)

Thus, frictional force is given by:

F = uN

F = umg cos(20)

Applying Work Energy theorem:

KE = -mg sin(20)*s - umg cos(20)*s

where s is the displacement along the ramp.

At the position when the block comes to a stop:

0 - 1/2mv^2 = -mg sin(20)*s - umg cos(20)*s

1/2v^2 = g sin(20)*s - ug cos(20)*s

1/2v^2 = g*s(sin(20) + u*cos(20)

1/2 x 1.24^2 = 9.81*0.14(0.342 + u*0.9397)

0.5598 - 0.342 = u*0.9397

**u = 0.251**

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