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

  • Thread starter Thread starter bjarnidk
  • Start date Start date
  • Tags Tags
    Sliding Slope
AI Thread Summary
An object sliding down a 30° slope at constant velocity indicates that the force of friction equals the component of gravitational force acting down the slope. The equations provided show that the frictional force can be expressed as μmg cos(30°) and must balance with mg sin(30°). The discussion highlights confusion over the calculation of the friction coefficient (μ) and the correct representation of forces involved. The correct frictional force is determined to be 5m Newtons, clarifying the earlier miscalculations. Understanding these relationships is crucial for accurately calculating energy loss on the slope.
bjarnidk
Messages
15
Reaction score
0

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...
 
Physics news on Phys.org
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
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top