Calculating Work Done on a Crate on an Incline

[teggenspiller]

Homework Statement

A n 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of 30 o with the horizontal. The force pushing the crate is parallel to the slope. If the speed of the crate increases at a rate of 1.5 m/s2, find the work done by the force:
A. 200 J
B. 1 J
C. 140 J
D. 200 J
E. 260 J

Homework Equations

The Attempt at a Solution

fnet=ma
fnet= (80/9.8)*(1.5)
12.24

w=12.24*5meters

w=61J


not an option!
aplease help

 

[cepheid]

The problem wants you to figure out the work done by the *applied* (pushing) force, not the work done by the net force.

 

[teggenspiller]

i think it is 200J after all

 

[gneill]

i think it is 200J after all

What is your reasoning? Can you show your calculations?

 

[supratim1]

yes right. 200J

 

[gneill]

yes right. 200J

Really?

The crate's on a 30° slope, so half of its 80N weight will be pressing downslope (since sine of 30° is 1/2). The net force to produce the desired acceleration was previously calculated to be 12.34N. So the upslope force required to yield this net force will be 12.34N + 40N, or 52.24N. How do you conclude that the work done is 200J, which is only 40N*5.0m?

 

[KaiserBrandon]

listen to Gneill, 200J is not the correct answer. Remember, you're trying to the find the work done by the force that is being applied to the crate. Determine what that force is and it should be straightforward from there.