Question

A 1.23-kg box slides up an incline plane with a speed of 8.6 m/s. If the angle of the plane with respect to the horizontal is 11 degrees and the coefficient of friction is 0.33, how long will it take for the box to come to rest?
A. 1.7 s
B. 2.5 s
C. 2.7 s
D. 17 s

Could you please explain how you arrive to the answer?

Answer 1

take the axes to be "tilted" along the incline. along the y-axis, Fg and Fn cancel out to 0. (there is no acceleration in this direction)

along an incline, Fg = mgcos(theta) in the y-direction. so Fn = mgcos(theta)

now, consider the x-axis. Fa is the applied force, Ff is friction. Also remember that gravity also acts along this axis. Friction force = μ(Fn) where μ is the coefficient of friction and Fn is the normal force we calculated from the first step. Force of gravity along the x-axis = mgsin(theta)

So net force in the x-direction: F_net = ma = μ(Fn) + mgsin(theta), plug in the appropriate quantities and solve for acceleration.

Finally, it's a simple kinematics problem with vf = vi + at where vf = 8.6 m/s, vi = 0, a is solved for in the previous step, and you can solve for t.