# Calculate acceleration and kinetic friction coefficient

1. Mar 19, 2016

### fission

1. The problem statement, all variables and given/known data

A particle of mass M is released from rest on a rough inclined plane, which makes an
angle of 30° with the horizontal. It is observed that the particle moves a distance of 3 m
in 3 s. What is the particle’s acceleration? Draw a properly labelled free body diagram.
Calculate the coefficient of kinetic friction between the particle and the surface of the
inclined plane.

2. Relevant equations

i made the equation of force with the help of formula F=mg. Equation i made was mg*sin 30 - kinetic friction force = ma

3. The attempt at a solution

i calculated coefficient of kinetic friction by tan 30 = 0.58
then i used equation mg.sin 30 - mg.cos30. 0.58 = ma.
m got divided on both sides but i am getting value of a -0.02 something so what does that mean? acceleration is negative? mass is going up instead of down? its very confusing please help me.

also how to use data it moves 3m in 3sec. why its given?

2. Mar 19, 2016

### haruspex

Every equation in physics has a context in which it is valid. What is the context for the equation $\mu=\tan \theta$?

3. Mar 19, 2016

### fission

its valid here by geometry. i cant show u by drawing a diagram here but if we draw a fbd of mass on inclined plane and forces acting on it then by equation $\mu= frictional force / normal force. we can have tan theta = cofficient of friction. 4. Mar 19, 2016 ### haruspex Only if frictional force/normal force equals tan theta. To get that, you are making an assumption that is not valid here. 5. Mar 19, 2016 ### fission then please help me how to solve this question? i am really stuck at it. 6. Mar 19, 2016 ### haruspex Write the ΣF=ma equations for two directions, either for the horizontal and vertical directions or for the tangent to plane and normal to plane directions. Your choice (but I recommend the second). 7. Mar 19, 2016 ### fission how can i do that? the body is not at rest and i cant balance any equilibrium forces. can you please provide me with a full solution. that would be really helpful. :) 8. Mar 19, 2016 ### haruspex You don't need the body to be at rest to use that equation. At rest would mean no acceleration, ΣF=0. But ΣF=ma is perfectly general. Homework Forum rules: we do not provide full solutions, just hints, verifications and corrections. 9. Mar 22, 2016 ### fission yea, i read the forum rules, sorry fr my ignorance. but the problem is i am not able to arrive at final solution. i tries to create 2 equations by taking the slope of inclined plane as x-axis and normal as y axis but still i wasn't getting adequate data to solve 2 variables. it still requires coefficient of friction value. idk where am i going wrong i have skipped this question for now from my assignment. can you at least please tell me what would be those two equations that would be a lot of help and dont we have to use the data 3m in 3s? Thanks a lot :) hope i get a solution this question really troubled me a lot xD 10. Mar 22, 2016 ### haruspex Don't just tell us you got equations you couldn't solve, post your working as far as you got. If we can't see what you did, we can't see where you are going wrong. 11. Apr 6, 2016 ### fission Hi, sorry for late reply. Was busy with my assignments. I did the solution this way in the pic below pls tell if its correct. I feel I can't use equation of motion here which I did. #### Attached Files: • ###### IMG_20160406_173452-1-1.jpg File size: 45.1 KB Views: 71 12. Apr 6, 2016 ### PeroK Looks good to me. 13. Apr 6, 2016 ### haruspex What you did was fine. What bothers you? 14. Apr 9, 2016 ### fission thx for verification, i was thinking i cant use equations of motion here as they are for 1-d motion but here we are working in 2-d plane. Also can u please tell me why do tan theta = coefficient of friction was not working here? what was i doing wrong there? thanks 15. Apr 9, 2016 ### PeroK If you have 2D motion, then you have the equations of motion separately in each direction. You could solve this problem by taking the horizontal and vertical directions as your two axes. But, that would be more difficult. What you did, therefore, was to take the direction down the slope as one axis and the normal direction as the other. This effectively reduced the problem to 1D motion. Why? Because the particle is constrained to move down the slope, so there is no motion in the normal direction. Why should$\mu = tan \theta$? That means friction would be dependent only on the angle. So, friction on an icy slope of angle$\theta$would be the same as friction on a rough slope of angle$\theta$. What you might be thinking is that the acceleration down a slope is related to$\mu$and$tan \theta##. You might like to try to derive that relationship.

16. Apr 9, 2016

### haruspex

The context in which that equation applies is constant speed. Most usually, it is applied where the particle is on the point of sliding, and it's μs.