Adisa pulls a 40-N crate up a 6.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 38° 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) 148 J
B) −148 J
C) 189 J
D) −189 J
E) None of these

Question

Adisa pulls a 40-N crate up a 6.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 38° 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) 148 J
B) −148 J
C) 189 J
D) −189 J
E) ​None of these

Quizplus · Accepted Answer

The net change in potential energy is calculated using the formula: PE = mgh, where m is mass, g is gravity, and h is height. The height (h) can be found using the sine of the incline angle: h = 6.0 m * sin(38°). The weight of the crate (40 N) acts as the force due to gravity. Thus, the change in potential energy is 40 N * 6.0 m * sin(38°) ≈ 148 J.

Adisa Pulls a 40-N Crate Up a 6