Question

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?

Answer 1

Answer:F=40.09 N

Explanation:

Given

weight of crate $W=50 N$

Inclination $\theta =37^{\circ}$

Frictional Force $f=10 N$

as the crate is moving with constant velocity therefore net Force on crate is zero

$F-50\sin (37)-f=0$

$F=50\sin (37)+10$

$F=30.09+10$

$F=40.09 N$