Work and Power Questions NEED Help

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Homework Help Overview

The discussion revolves around a physics problem involving work and power, specifically focusing on a scenario where a crate is pulled along a horizontal surface with friction. The problem requires determining the work done by various forces acting on the crate.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the components of the pulling force and question how to calculate the work done by each force, including friction. There is an attempt to resolve the force into horizontal and vertical components, and discussions arise about the implications of vertical movement on work done.

Discussion Status

Participants have engaged in resolving the forces and calculating work done in the horizontal direction. Some have provided numerical values for the components of the force and the work done, while others are clarifying the conditions under which work is done vertically. The conversation reflects a mix of attempts and confirmations without reaching a consensus on the overall solution.

Contextual Notes

There is an emphasis on understanding the setup of the problem, including the effects of friction and the angle of the applied force. Participants are also considering the implications of the crate's vertical movement on the work done.

Turnips
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1. A 50kg crate is pulled 40m along a horizontal floor by a constant force exerted by a person, Fp=100N, which acts at a 37 degree angle. The floor is rough and exerts a friction force Ff= 50N. Determine the work done by each force acting on the crate (including normal force and force of weight), and the net work done on the crate.



W=F*d*cos( angle )

3. no idea how to do it.
 
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What is the magnitude of the component of Fp acting in the direction of motion?
 
Im guessing its just 100N because that's all the problem gives me.
 
Are you familiar with http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l3b.html ?
 
Last edited by a moderator:
Yes.
 
Then resolve the 100N acting at 37 degrees into horizontal and vertical components.
 
The Fy = 60.18N and the Fx = 79.86N
 
Turnips said:
The Fy = 60.18N and the Fx = 79.86N
Good.

Now what would be the work done in moving the crate 40 m in the direction of Fx? and in the same manner what is the work done by the frictional force?
 
79.86N * 40 m = 3194.4J is the Work done in the x Direction
50N * 40m = 2000J is the work done by Friction
 
  • #10
Turnips said:
79.86N * 40 m = 3194.4J is the Work done in the x Direction
50N * 40m = 2000J is the work done by Friction

That is what the question asks for.

Now vertically. Is the crate moving either up or down? If so then some force is doing work on it. If not then no work is done on the crate
 
  • #11
Alright thanks very much.
 

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