Slanted surfaces with friction: A box is sliding down an incline tilted at a 12.0° angle above horizontal. The box is initially sliding down the incline at a speed of 1.50 m/s. The coefficient of kinetic friction between the box and the incline is 0.340. How far does the box slide down the incline before coming to rest?
A) 0.920 m
B) 2.33 m
C) 1.78 m
D) 0.720 m
E) The box does not stop. It accelerates down the plane.

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the following equation of motion for an object sliding down an inclined plane with friction:

v^2 = u^2 + 2as - 2ukgsinθ

where v is the final velocity (0 m/s, since the box comes to rest), u is the initial velocity (1.50 m/s), a is the acceleration (unknown), s is the distance traveled (unknown), uk is the coefficient of kinetic friction (0.340), g is the acceleration due to gravity (9.81 m/s^2), and θ is the angle of the incline (12.0°).

We can rearrange the equation and solve for s:

s = (v^2 - u^2) / 2a + ukg sinθ

s = (0 - (1.50)^2) / (2(-ukg sinθ))

s = 0.920 m

Therefore, the box slides down the incline for a distance of 0.920 m before coming to rest.

Slanted Surfaces with Friction: a Box Is Sliding Down an Incline