Work Done by Weight Homework: Find W

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SUMMARY

The discussion centers on calculating the work done by weight on a 700 kg crate on a 30° incline, subjected to a horizontal force of 5600 N. The participant initially misinterprets "work done by weight" as the work done by the applied force P but later clarifies it refers to the work done by gravity (ΔPE). The conversation evolves to include finding the work done by friction when the velocity changes from 2.5 m/s to 2.3 m/s over a distance of 3.0 m, requiring the application of the conservation of energy principle.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of potential energy (PE) and kinetic energy (KE)
  • Ability to interpret free-body diagrams
  • Knowledge of the work-energy principle
NEXT STEPS
  • Learn how to calculate work done by gravitational forces in inclined planes
  • Study the work-energy theorem and its applications in physics problems
  • Explore frictional force calculations and their impact on motion
  • Investigate the conservation of energy in mechanical systems
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to work, energy, and forces on inclined surfaces.

Legerity
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Homework Statement



A 700 kg crate is on a rough surface inclined at 30°. A constant external force P = 5600 N is applied horizontally to the crate. The force pushes the crate a distance of 3.0 m up the incline, in a time interval of 7.3 s, and the velocity changes from v1 = 1.4 m/s to v2 = 2.5 m/s. What is the work done by weight?

75512f21-629f-400f-914b-f9f928f9ba49.png


Homework Equations



W = Fd, KE = 0.5mv^2, PE = mgh

The Attempt at a Solution



I drew the free-body diagram, but I am stumped on how to go about the problem. What is the work done by weight, and how can I find it? I don't exactly know what it is.
 
Last edited:
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I don't know what they mean by 'work done by weight'. My guess is they meant the work done by the force P. (Cut and paste error from another question?)
 
haruspex said:
I don't know what they mean by 'work done by weight'. My guess is they meant the work done by the force P. (Cut and paste error from another question?)
I figured out the problem, and it is the work done by gravity on the box (ΔPE).
 
Legerity said:
I figured out the problem, and it is the work done by gravity on the box (ΔPE).
Really? So most of the information is irrelevant? (Or maybe there are more parts to the question?)
 
haruspex said:
Really? So most of the information is irrelevant? (Or maybe there are more parts to the question?)
Yes, there is another part of the question I am stuck on. It is now asking me to find the work done by the frictional force, and my attempts at the problem have been futile. The only things changed in the info in the original post is that now V2 = 2.3 m/s, and time = 8.3 s. Could you possibly help?
 
Last edited:
You'll have to use an equation that relates conservation of energy and the energy lost by friction: KE_{i}+PE_{i}+WE_{i}=KE_{f}+PE_{f}+WE_{f}
 
Legerity said:
It is now asking me to find the work done by the frictional force,
You know the distance, so it remains to calculate the frictional force. What equations do you get from your free body diagram?
 
Legerity said:
I figured out the problem, and it is the work done by gravity on the box ( - [/color]ΔPE).
see important correction in red[/color].
 

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