# Crate pulled up rough incline

## Homework Statement

A crate is pulled up a rough incline. The pulling force is parallel to the incline. The crate is pulled a distance of 5.79 m. The acceleration of gravity is 9.8 m/s^2
theta=23.4 degrees
mass= 8.91kg
Force pulling on box= 142N
coefficient of friction= 0.292
Initial velocity of box= 1.48m/s

A.What is the magnitude of the work is done by the gravitational force? Answer in units of J.
B.How much work is done by the 142 N force? Answer in units of J.

## Homework Equations

PE= mgh
W=F*displacement*cos(theta)

## The Attempt at a Solution

A. I tried using the mgh formula but height is not given. Then, I tried the second formula given but I do not know how to calculate the displacement.
B. Used the second equation got W= 142N*5.79*cos(23.4). Ended up being incorrect.
UPDATE: got the second one as the cos(theta) ended up being 1 as the angle between the displacement and the force was 0.

Clearly lost. I do not know where to properly start so a hint at that would be helpful![/B]

Last edited:

hilbert2
Gold Member
Calculating the height from the distance is basic trigonometry. The pulling force of 142 N has the same direction as the displacement, so you don't need any trig functions in part B.

Calculating the height from the distance is basic trigonometry. The pulling force of 142 N has the same direction as the displacement, so you don't need any trig functions in part B.
So cos(23.4)=x/5.79. 5.79cos(23.4)=x. x=5.314
mgh= 8.91*9.8*5.314= 463.99
Does this mean the work done by gravity is 463.99?

hilbert2
Gold Member
^ If theta is the angle between the incline and the horizontal direction, then h = d sin(theta), where d = 5.79m

^ If theta is the angle between the incline and the horizontal direction, then h = d sin(theta), where d = 5.79m
Plugging in I get h= 2.299m
mgh=8.91*9.8*2.299= 200.786
Does this mean the work done by gravity is 200.786?

hilbert2