Question

A cube of mass M=0.703 kg is placed on an incline and given a push. It slides down, and its speed down the incline is measured at two positions, [1] and [2] on the diagram below, with V1 = 0.923 m/s and V2 = 1.207 m/s. Note that the incline passes through at least two grid intersections and that one cube corner is at x grid lines.

A. Calculate mk, the coefficient of kinetic friction for the block on the incline.



The height of the Right triangle is .38m and the length is 1m

Answer 1

Help will be provided on problems where you have shown work. This is not one of them.

Answer 2

Ok here is where I have got so far.
Tan (theta) = .38m/1m
theta = 20.807 deg

Sum Fy=mg*sin(theta)-Fn=ma=0
.703kg(9.81)*sin(20.807)-Fn=o
Fn=2.45N

Sum Fx=mgcos(theta)-Ff=ma
.703(9.81)*cos(20.807)-mk(2.45)=.703*a

This is where I am stuck

Answer 3

Thank you for showing your work!

The angle is arctan0.38 = 20.81 degrees, as you calculated.

Unfortunately, I need to know the vertical (H) or horizontal distance between the two measurement stations in order to proceed. I don't know what you mean by "grid intersection" or "cube corner". This would probably make sense if I could see the figure.

The decrease in potential energy between the two measurement stations, M g H, equals the kinetic energy increase,
(M/2)(V2^2 - V1^2)
PLUS
the work done against friction,
M g cos20.8 *mk* H/sin20.8
= M*g*H*mk/tan20.8

M cancels out. Solve for mk