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 $\degree$ 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?
A)120 J
B)-120 J
C)200 J
D)-200 J

Question

Quizplus · Accepted Answer

The net change in potential energy is calculated using the formula: PE = mgh. The height (h) can be found using the sine of the incline angle: h = 5.0 m * sin(37°). The weight (mg) is 40 N. Thus, PE = 40 N * 5.0 m * sin(37°) = 120 J.

Adisa Pulls a 40-N Crate Up a 5 °\degree° To the Horizontal and There Is a Constant Force of Plane

Adisa Pulls a 40-N Crate Up a 5 $\degree$ To the Horizontal and There Is a Constant Force of Plane