A box is sliding down an incline tilted at a 11.1° angle above the horizontal. The box is initially sliding down the incline at a speed of 1.70 m/s. The coefficient of kinetic friction between the box and the incline is 0.390. How far does the box slide down the incline before coming to rest.
a) 1.08 m
e)The box does not stop. It accelerates down the plane
U1 + K1 + Wother = U2 + K2
The Attempt at a Solution
Given an initial velocity and a coefficient of kinetic friction, immediately I thought of using the work energy theorem to solve this problem.
I first set up a coordinate system that is tilted and running along the plane, with the origin at the top of plane.
I assumed the block starts from the top (with given initial velocity) and ends at a distance d - the unknown variable we're solving for.
I drew a free body diagram of the box and summed the forces in both the X an Y directions.
Here is the result of that:
ƩFx= mgsinθ - Ffriction
I ignored the Fy equation and used the Fx one to get:
so now I set up my Work-Energy Theorem Equation exactly like how I wrote it in part 2 above:
0 + (1/2)mv2 - μmgcosθd = 0 + 0
The equation is set to zero because the block stops at the end of the interval (∴ K2= 0) and there is never potential energy as per my coordinate system (∴U2=0)
I solved my Work-Energy equation for 'd' and got 0.3852819047≈0.385.
Can anyone tell me where I went wrong? Did I take the wrong approach? which one should I take?