Question

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?

Answer 1

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

$PE = mgh$

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

$h=Lsin\theta$

$h = 5*sin37$

$h = 3m$

Applying the potential energy formula we have to

$PE = mgh$

$PE = F_g(h)$

$PE = 40N*3m$

$PE = 120J$

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