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

Step-by-step 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`