A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to
the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker?
1 answer:
Answer:F=40.09 N
Explanation:
Given
weight of crate 
Inclination 
Frictional Force 
as the crate is moving with constant velocity therefore net Force on crate is zero




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The answer is 0.981 J
E = m · g · h<span>
E - energy
m - mass
g - gravitational acceleration
h - height
We know:
E = ?
m = 0.10 kg
g = 9.81 m/s</span>²
h = 1 m
E = 0.10 kg * 9.81 m/s² * 1 m = 0.981 J
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