Kinetic friction and lost mechanical energy

Click For Summary
SUMMARY

The discussion focuses on calculating kinetic friction and the mechanical energy lost by an object sliding down a 30-degree incline. The friction coefficient (µ) is determined using the equation F(friction) = µmgcos(θ) = mgsin(θ), leading to the conclusion that µ = tan(θ). The participants clarify that the mechanical energy lost due to friction is equal to the thermal energy generated, and they emphasize the importance of distinguishing between force and energy units. The rate of energy dissipation is linked to the friction force and potential energy loss per unit time, although the exact velocity of the object is not provided.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of kinetic and potential energy
  • Knowledge of friction coefficients and their calculations
  • Basic grasp of energy conservation principles
NEXT STEPS
  • Calculate kinetic friction using the formula F(friction) = µmgcos(θ)
  • Explore energy conservation in mechanical systems, focusing on potential and kinetic energy
  • Investigate the relationship between friction force and energy dissipation over time
  • Study the implications of constant velocity on energy loss in inclined planes
USEFUL FOR

Students and educators in physics, mechanical engineers, and anyone interested in understanding the dynamics of friction and energy loss in inclined motion.

hunbogi
Messages
6
Reaction score
0
An Object slides with a consistent velocity down an incline that has a 30 degree angle.
I'm trying to find the kinetic friction and the ratio that the mechanical energy is lost on the way down.

I think I have already calculated µ with F(friction)= µmgcos(θ)=mgsin(θ) out comes Tan(θ) = µ
I think I did that correctly..
I'm struggling with the other part about the rate that the mechanical energy is lost. I would guess that it is not possible to get the answer in a number.
3.
I tried to us PE = mgh -friction = k2.
mgh - µmghcos(θ) = ½mv^2þ
not sure what my answer might look like in the end.
 
Physics news on Phys.org
I agree with you that u= tan(theta). As for the ratio- the ratio of mechanical energy lost to what? Thermal energy maybe?
 
It only says find the mechanical energy that is lost because of friction on the way down the incline.
I think U =mgs is the potential energy. energy on the end of the incline is K = ½mv^2
so I think the diffrent between the two is the friction force.
 
Ok so there is no ratio. I think your getting force and energy confused also. Be aware that force has units of Newtons= kg*m/s^2 and energy has units of Joules= N*m= kg*m^2/s^2. The mechanical energy lost is equal to the thermal energy generated. So if T is thermal energy then using conservation of energy set ΔKE+ΔPE+T= 0 and solve for T. btw this should probably not be posted in advanced physics
 
Okey thanks I will try that. New here so I didn't know where to post
 
Well, hello hunbogi, and welcome to PF :)

I suppose the consistent velocity is a constant velocity and the ratio is the rate, i.e. the mechanical energy loss per unit time.
There are two ways to find that rate:
One is friction force times displacement per unit time
Two is from potential energy loss per unit time
Answers should be the same
 
The rate of energy dissipation will depend on the velocity. Is the velocity given?
 
no it only gives the angle 30 degree and says the velocity is consistent
 
Consistent velocity does not make much sense. Is this a translation from another language?
If it's constant velocity, it can have any value, including zero. In which case there is no energy dissipation.
Something is missing and it may be due to translation.
 
  • #10
Sorry it is constant velocity. mixed up in translation
 
  • #11
Can you post the entire problem, as it is? Not just a summary.
 
  • #12
3.
An object slides with a constant velocity down an 30 degree incline
a) Find the friction force and find the ratio of the mechanical energy that is lost on the way down
 
  • #13
Oh, I think you want to say "the fraction of mechanical energy". This could make sense.
It may ask "What fraction of the mechanical energy is lost due to friction". Like, one half or 1/3 or 0.2.

Still does not seem completely alright. It is not clear what mechanical energy they are talking about.
The kinetic energy does not change. So the decrease in potential energy will be equal to the work of the friction force.
 

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
4K
Replies
2
Views
1K
Replies
1
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
2
Views
2K