How Do You Calculate Friction and Energy Loss on a Slope?

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Homework Help Overview

The problem involves an object sliding down a slope inclined at 30° to the horizontal, with a focus on calculating the force of friction and energy loss during the descent while maintaining constant velocity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the balance of forces acting on the object, particularly the relationship between the force of friction and the gravitational components along the slope. Questions arise regarding the representation of friction in the equations and the accuracy of calculated values.

Discussion Status

Several participants are actively engaging with the problem, questioning the calculations presented and exploring different representations of the force of friction. There is a focus on clarifying the relationships between the variables involved, though no consensus has been reached on the correct values or methods.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is an emphasis on understanding the underlying physics rather than simply obtaining a numerical answer.

bjarnidk
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Homework Statement


An object slides down a slope of 30° to the horizontal with CONSTANT VELOCITY.
Find the force of friction and the energy lost on the way down.

Homework Equations


[tex]\mu m g \cos \theta[/tex]
[tex]\mu m g \cos \theta - m g \sin \theta = 0[/tex]
[tex]W = K_2 + U_2 - (K_1 + U_1)[/tex]

The Attempt at a Solution


I tried the following:

[tex]\mu mg \cos30 = mg \sin30 = \mu = tan30 \cdot m = 3,3 \cdot m[/tex]

So the ratio is [tex]3,3m[/tex]? It doesn't sound right...
 
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In your force balance equation, the two terms are equal to one another, so they must both be representations of the force of friction (Actually, the second term is the component of the body's weight along the incline, which, as you showed, is equal to the force of friction). Which of these two representations of the force of friction do you think will be easier to work with in the second part of your problem, given that one of them contains μ, and the other term contains only quantities that you know in advance?

Chet
 
Doesn't look right either! You divide left and right by cos 30, which is good. Why then divide by m only on the left and not on the right ?

Then: tan 30° is definitely not 3.3
 
Sorry, [tex]\mu[/tex] is tan30, but the force of friction is 3,3m, would that be correct?
 
bjarnidk said:
Sorry, [tex]\mu[/tex] is tan30, but the force of friction is 3,3m, would that be correct?

No. As I said in my previous post, the friction force is mgsin30=5m Newtons
 

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