- #1
Smith96
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Homework Statement
The first problem was " A 50 N crate is pulled up a 5m inclined plan by a worker at constant velocity. If the plane is inclined at an angle of 37degrees to the horizontal and there exists a constant frictional force of 10N between the crate and the surface, what is the force applied by the worker?"
I knew that I could find the frictional force coefficient by 10=(coefficient)(mass)(gravity) =>
10=(coefficient)(50N)(9.8) => coefficient =.02
and then using the equation (coefficient)(mass)(gravity)(distance) I could get the force applied.
The next problem was "A 40 N crate is pulled up a 5m inclined plan by a worker at constant velocity. If the plane is inclined at an angle of 37degrees to the horizontal and there exists a constant frictional force of 10N between the crate and the surface, what is the net change in potential energy?"
I thought I would use (mass)(distance).
With the first problem the distance would be cos(37) and in the second its sin(37)
Why is that?
Homework Equations
(friction coefficient)(mass)(gravity)(distance)
The Attempt at a Solution
For the first, (50)(5cos(37))(9.8)(.02)=40N
For the second, (5sin37)(40N)=120J