A dock worker gives a crate a strong push up a slippery 20° ramp (μk =...

A dock worker gives a crate a strong push up a slippery 20° ramp (μk = 0.1). The crate elevates 2.6 m higher than its initial y position at the release point of the worker's hands before coming to a stop on the ramp. Determine the speed at which the crate was moving when released.

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Expert Solution

Gravitational acceleration = g = 9.81 m/s2

Mass of the crate = m

Height of the ramp = H = 2.6 m

Angle the ramp is inclined at = = 20 degrees

Length of the ramp = L

L = 7.602 m

Normal force on the crate from the ramp = N

Coefficient of kinetic friction between the crate and the ramp = k = 0.1

Friction force on the crate = f

Initial speed of the crate = V

By conservation of energy the initial kinetic energy of the crate is equal to the potential energy gained by the crate plus the work done against friction as the crate rises up the ramp.

V = 8.06 m/s

Speed of the crate when it was released = 8.06 m/s

genius_generous answered 2 years ago

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