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Adisa pulls a 40-N crate up a 5.0-m long inclined plane at a constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there is a constant force of friction of 10 N between the crate and the surface, what is the net change in potential energy of the crate?

Respuesta :

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To solve this problem we resort to the concept of potential energy which is given by the equation

[tex]PE = mgh[/tex]

Where,

m= mass

g= Gravitational acceleration

h = Height

The height is given in the form of a component, that is, it is given the length that is 5 m and the angle of 37 degrees, therefore the height would be

[tex]h=Lsin\theta[/tex]

[tex]h = 5*sin37[/tex]

[tex]h = 3m[/tex]

Applying the potential energy formula we have to

[tex]PE = mgh[/tex]

[tex]PE = F_g(h)[/tex]

[tex]PE = 40N*3m[/tex]

[tex]PE = 120J[/tex]

Therefore the net change in potential energy of the crate is 120J.

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