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

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SUMMARY

The discussion focuses on calculating the force of friction and energy loss for an object sliding down a 30° slope with constant velocity. The key equations involved are μmg cos θ and mg sin θ, which represent the force of friction and the gravitational force component along the incline, respectively. The correct force of friction is established as 5m Newtons, derived from the equation mg sin 30°. The participants clarify the relationship between the coefficient of friction (μ) and the forces acting on the object.

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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


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

The Attempt at a Solution


I tried the following:

\mu mg \cos30 = mg \sin30 = \mu = tan30 \cdot m = 3,3 \cdot m

So the ratio is 3,3m? 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, \mu is tan30, but the force of friction is 3,3m, would that be correct?
 
bjarnidk said:
Sorry, \mu 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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