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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?

Respuesta :

Answer:F=40.09 N

Explanation:

Given

weight of crate [tex]W=50 N[/tex]

Inclination [tex]\theta =37^{\circ}[/tex]

Frictional Force [tex]f=10 N[/tex]

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

[tex]F-50\sin (37)-f=0[/tex]

[tex]F=50\sin (37)+10[/tex]

[tex]F=30.09+10[/tex]

[tex]F=40.09 N[/tex]

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