Asked by Misaki
A 13.2 kg crate is pulled by a 157 N force (parallel to the incline) up a rough incline. Assume that the incline makes an angle of 34.0° with the horizontal and that the coefficient of kinetic friction is 0.434. The crate has an initial speed of 1.53 m/s. It is pulled a distance of 6.30 m. What is the change in gravitational potential energy of the crate? Answer in units of J.
I thought that the work done by the force of gravity would be equal to the change in gravitational potential energy, (-465), but it turned out to be wrong. Any suggestions?
I thought that the work done by the force of gravity would be equal to the change in gravitational potential energy, (-465), but it turned out to be wrong. Any suggestions?
Answers
Answered by
Damon
6.3 * sin 34 = 3.52 meters up
m g h = 13.2 * 9.81 * 3.52 = 456
so I get
-456 J
I think you reversed digits
m g h = 13.2 * 9.81 * 3.52 = 456
so I get
-456 J
I think you reversed digits
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