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