Push a 22kg crate up a frictinless incline

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Homework Help Overview

The discussion revolves around calculating the work done on a crate being pushed up a frictionless incline. The crate has a mass of 33 kg and is subjected to various forces, including an applied force and gravitational force, while moving along an incline angled at 15°.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of work done by different forces acting on the crate, including the applied force, gravitational force, and normal force. There is uncertainty about how to combine these individual works to find the total work done.

Discussion Status

Some participants have shared their calculations for the work done by individual forces, while others express confusion about how to derive the total work from these values. There is a recognition of the need to consider all forces acting on the crate in the total work calculation.

Contextual Notes

Participants note that the problem involves multiple forces and their contributions to the work done, with specific values already calculated for individual components. There is an emphasis on understanding how these forces interact rather than simply applying formulas.

EmoryGirl
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Homework Statement


To push a 33 kg crate up a frictionless incline, angled at 15° to the horizontal, a worker exerts a force of 100.4 N, parallel to the incline. As the crate slides 1.50 m, (a)how much work is done on the crate by the worker's applied force? (b) How much work is done on the crate by the weight of the crate? (c) How much work is done on the crate by the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?


Homework Equations



I can NOT solve (d)!

The Attempt at a Solution


I have solved a, b, and c:
(a) 1.51 x 10^2 J
(b) -1.26 x 10^2 J
(c) 0.00 J
(d) ?
 
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Hi EmoryGirl,

EmoryGirl said:

Homework Statement


To push a 33 kg crate up a frictionless incline, angled at 15° to the horizontal, a worker exerts a force of 100.4 N, parallel to the incline. As the crate slides 1.50 m, (a)how much work is done on the crate by the worker's applied force? (b) How much work is done on the crate by the weight of the crate? (c) How much work is done on the crate by the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?


Homework Equations



I can NOT solve (d)!

The Attempt at a Solution


I have solved a, b, and c:
(a) 1.51 x 10^2 J
(b) -1.26 x 10^2 J
(c) 0.00 J
(d) ?


The total work done on the crate is the work done by all the forces that are acting on the crate. What would that be?
 
alphysicist said:
Hi EmoryGirl,




The total work done on the crate is the work done by all the forces that are acting on the crate. What would that be?

The forces acting on the crate include the applied force and mg. I'm not sure how they work together in an equation...
I tried just W=Fd using 100.4N as the the F but I know that is not right
 
EmoryGirl said:
The forces acting on the crate include the applied force and mg. I'm not sure how they work together in an equation...
I tried just W=Fd using 100.4N as the the F but I know that is not right

It's not right because it is only the work from the applied force. Remember that you already have the work done from each individual force that is acting on the crate; what then is the total work done?
 
Okay I just solved it! I added the work done by the applied force to the work done by the weight of the crate. The answer is 25 J...I didn't realize that it was so simple!
Thanks for your help!
 
EmoryGirl said:
Okay I just solved it! I added the work done by the applied force to the work done by the weight of the crate. The answer is 25 J...I didn't realize that it was so simple!
Thanks for your help!

Glad to help, and welcome to PhysicsForums!
 

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