Calculating Work Done on a Crate on an Incline

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Homework Help Overview

The discussion revolves around calculating the work done on an 80-N crate being pushed up a 30° incline over a distance of 5.0 m, with a given acceleration of 1.5 m/s². Participants are exploring the implications of different forces acting on the crate and the correct interpretation of the problem statement.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the distinction between the work done by the applied force versus the net force. There are attempts to calculate the required force and the corresponding work, with some participants expressing confidence in their answers while others question the reasoning behind those conclusions.

Discussion Status

The conversation is ongoing, with some participants suggesting that the answer is 200 J while others challenge this conclusion by emphasizing the need to consider the applied force correctly. There is a mix of agreement and skepticism regarding the calculations presented.

Contextual Notes

Participants are grappling with the definitions of forces involved and the specific requirements of the problem, including the need to calculate work done by the applied force rather than the net force. There is also a mention of the sine of the incline angle affecting the weight component acting along the slope.

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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
 
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The problem wants you to figure out the work done by the *applied* (pushing) force, not the work done by the net force.
 
i think it is 200J after all
 
teggenspiller said:
i think it is 200J after all

What is your reasoning? Can you show your calculations?
 
yes right. 200J
 
supratim1 said:
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?
 
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.
 

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