Question

A brick of mass 1.7 kg was pushed down a roof with an initial velocity of 0.4 m/s. The roof is inclined at 21.9^o with respect to the horizontal, and the coefficient of dynamic friction between the roof and the brick is 0.7. The edge of the roof is 3.9 m from where the brick was pushed.

How fast will the brick be moving when it reaches the edge of the roof (in m/s)?

Or, perhaps, the brick will not reach the edge of the roof? In that case, how far from the edge will it stop (in m)?

Answer 1

Work done by all forces = change in KE,

mgL sin theta - umg L cos theta = 0.5m*(v^2-u^2)

1.7*9.8*3.9* sin 21.9 degree - 0.7*1.7*9.8*3.9* cos 21.9 degree = 0.5*1.7*(v^2 - 0.4^2)

The equation does not have any real root, it means the brick didn't reach the edge,

Let it reach x from edge,

Work done by all forces = change in KE,

mgL sin theta - umg L cos theta = 0.5m*(v^2-u^2)

1.7*9.8*(3.9-x)* sin 21.9 degree - 0.7*1.7*9.8*(3.9-x)* cos 21.9 degree = 0.5*1.7*(0 - 0.4^2)

x = 3.87 m answer