Question

a 25 kg box is released from rest on a rough inclined plane tilted at an angle of 33.5 degrees to the horizontal. The coefficient of kinetic friction between the box and the inclined plane is 0.200.

a. determine the force of kinetic friction acting on the box
b. determine the acceleration of the box as it slides down the inclined plane.

Answer 1

Normal force = m g cos 33.5
friction force = .2 m g cos 33.5

weight force component down slope = m g sin 33.5

total force down slope= mg(sin33.5-.2cos33.5)

a = F/m =g(sin33.5-.2cos33.5)

Answer 2

H2(g)+N2(g)NH3(g) Please help balance this equation. Thank you