Solving Incline Problem with Normal Force: 39.36

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Homework Help Overview

The discussion revolves around a physics problem involving an object sliding up a frictionless incline. Participants are exploring the concepts of gravitational force, normal force, and work done against gravity in the context of the incline's angle.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the use of gravitational force and normal force, questioning the assumption that gravitational force is zero on an incline. They explore the angle to use in calculations and whether potential energy differences can be applied.

Discussion Status

The discussion is ongoing, with participants sharing various attempts to solve the problem and questioning each other's reasoning. Some guidance has been offered regarding the angles and the signs in calculations, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the methods they can use. There is also a focus on understanding the relationship between forces and motion on an incline.

momoneedsphysicshelp
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Homework Statement
A 4.9 kg mass slides 3.6 m up a frictionless 35 degree incline. How much work is done by gravity?

I keep getting this question incorrect and I don't know what I'm doing wrong.
Relevant Equations
W= F*d*cos(theta)
first I tried to use gravitational force to solve it but because of the incline the gravitational force is zero. so I used normal force to solve.
Normal force is 39.36,
then f*d*cos(theta)
39.36*3.6*cos(55)
it would be cos 55 because the angle is in the direction of normal force from the position of the object
it would equal= 39.36*3.6*cos(55)= 81.27
but it is incorrect.

can anyone please tell me where I made a mistake or if there is another way I should be solving this problem?
thanks
 
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Why would the gravity be zero on the incline? Gravity is always present. Being on an incline doesn’t change that.
 
Are you allowed to do it by potential energy difference?
 
neilparker62 said:
The angle you need to use is that between the gravitational force and the direction of motion.
I tried that but it was still incorrect
 
momoneedsphysicshelp said:
I tried that but it was still incorrect
Did you take the sign into account? Remember that gravity is working against the direction of motion here.
 
I deleted because I saw you did that. But maybe it's just the sign which is wrong since work is done against gravity rather than by gravity.
 
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momoneedsphysicshelp said:
I tried that but it was still incorrect
So what angle did you use? And what answer did you get?
 
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suremarc said:
Did you take the sign into account? Remember that gravity is working against the direction of motion here.
Yes, it was still incorrect
 
  • #10
haruspex said:
So what angle did you use? And what answer did you get?
The angle I used was the angle in between the direction of gravity and the object itself, but I tried the complementary angle to that and it was still incorrect.
 
  • #11
You've tried several things, none of which gave you the correct answer. Can we see your best attempt?
 
  • #12
Here are some questions to consider:
a) When a 4.9kg mass falls 1m, how much work is done by gravity?
b) When a 4.9kg mass rises by 1m, how much work is done by gravity?
c) If an object slides a distance 3.6m up a 35º incline, what is its change in height?
 
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  • #13
momoneedsphysicshelp said:
I used normal force to solve.
An object on an incline does not move in the direction of the normal force. It follows that the normal force does no work on the object. Nor does the normal component of the gravitational force.
 
  • #14
momoneedsphysicshelp said:
Homework Statement:: A 4.9 kg mass slides 3.6 m up a frictionless 35 degree incline. How much work is done by gravity?

I keep getting this question incorrect and I don't know what I'm doing wrong.
Relevant Equations:: W= F*d*cos(theta)

first I tried to use gravitational force to solve it but because of the incline the gravitational force is zero. so I used normal force to solve.
Normal force is 39.36,
then f*d*cos(theta)
39.36*3.6*cos(55)
it would be cos 55 because the angle is in the direction of normal force from the position of the object
it would equal= 39.36*3.6*cos(55)= 81.27
but it is incorrect.

can anyone please tell me where I made a mistake or if there is another way I should be solving this problem?
thanks

Thank you all fro the help, My mistake was using cosine instead of sin and not considering that gravity would be negative
 

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