Kinetic Energy and Work Homework Help

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Homework Help Overview

The discussion revolves around a physics problem involving a 50 kg mass being pulled up a 20° incline. The problem requires participants to calculate the work done by tension, friction, and against gravity, while considering the effects of kinetic friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants express confusion regarding the setup of the problem and the application of the work formula, specifically the variables involved in Fd x cosθ. There are attempts to clarify the role of free body diagrams (FBD) in visualizing forces. Some participants discuss calculating forces such as gravitational force and normal force, while questioning how to derive the frictional force.

Discussion Status

There is ongoing exploration of the problem, with participants sharing their understanding of forces and equations. Some guidance has been offered regarding the use of free body diagrams and the relationships between forces, but no consensus has been reached on the specific calculations needed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is an emphasis on understanding the forces involved and their respective directions.

KTiaam
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Homework Statement


A 50 kg mass is pulled at a constant speed up a 20° incline which is 12m long and 4.1m high.

μk = 0.32

a) find the work done by tension, T.
b) find the work done by friction.
c) find the work done against gravity.


Homework Equations



W = F x D
Work = force x Distance

W = Fd x cosθ


The Attempt at a Solution



I still confused on the concept of this.

i know that you have to use Fd x cosθ, but i am confused on how to set it up.
 
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KTiaam said:

Homework Statement


A 50 kg mass is pulled at a constant speed up a 20° incline which is 12m long and 4.1m high.

μk = 0.32

a) find the work done by tension, T.
b) find the work done by friction.
c) find the work done against gravity.


Homework Equations



W = F x D
Work = force x Distance

W = Fd x cosθ


The Attempt at a Solution



I still confused on the concept of this.

i know that you have to use Fd x cosθ, but i am confused on how to set it up.

Always start this kind of problem with a free body diagram (FBD). Can you sketch one and post it as an attachment? And use the FBD to show the forces that you are being asked about in the problem...
 
Frictional force is the constant of friction multiplied by the normal force, which is the force equal and opposite to the gravitational force.
 
berkeman said:
Always start this kind of problem with a free body diagram (FBD). Can you sketch one and post it as an attachment? And use the FBD to show the forces that you are being asked about in the problem...

This is my FBD
 

Attachments

  • photo.JPG
    photo.JPG
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KTiaam said:
This is my FBD

Good, so now you can start writing equations for each of the works that you are asked to find...
 
berkeman said:
Good, so now you can start writing equations for each of the works that you are asked to find...

Im still confused on how to do that, all i know is that:

f// = mgsinθ
fn = mgcosθ
 
KTiaam said:
i know that you have to use Fd x cosθ, but i am confused on how to set it up.
What specifically is confusing you about applying the formula? You have three quantities, F, d, and θ. What do these variables represent?
 
KTiaam - remember to put the frictional force in your free-body diagram.
 
So i know how to find F of gravity and force perpendicular how do i find force friction?

Since i know F perpendicular would it be just the opposite?

F perpendicular is equal to 46.045
i got that by:

mass x gravity x cos 20°

= 5 kg x 9.8 x cos 20°
=46.045

would force normal be negative then?
or is force of gravity negative?

i know that Force of friction is:

coefficient of friction x force normal

am i on the right track?
 
  • #10
i know that Force of friction is:

coefficient of friction x force normal

Yes.
For the direction of the frictional force - remember that it always opposes the motion.
 

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