How to work out friction coefficient of mass which is decelerating.

So we got

Particle A is going down a slope of 14m. At the start, it was at 12 ms-1 and at the end it was 8 ms-1.

The mass of A is 3.5kg, and the slope is angled at 30 degrees.

Work out the friction coefficient.

How would I do this?????

So we got

Particle A is going down a slope of 14m. At the start, it was at 12 ms-1 and at the end it was 8 ms-1.

The mass of A is 3.5kg, and the slope is angled at 30 degrees.

Work out the friction coefficient.

How would I do this?????

I'd probably use energy. Have you learnt the work/energy principle?

yes. That was the previous question.

I worked out the work done as 140J

Can you please post what you've tried for this question?

yes

Energy lost = work done = change in Kinetic Energy

Change in kinetic energy = (1/2 x 3.5 x 12^2) - (1/2 x 3.5 x 8^2)
which gave me 140J

I don't see now how I can use this. I can work out the force of the particle going down the slope, which would be 140/14 (since 14 is the displacement, and using the formula work done = force x displacement)

Then this means that Fmax would be 10, since it's decelerating, meaning that friction is at it's maximum? Is it right doe? This is the part where I get stuck.

Then this means that Fmax would be 10, since it's decelerating, meaning that friction is at it's maximum? Is it right doe? This is the part where I get stuck.

You haven't considered the gain in gravitational potential energy:

Energy lost = Loss in KE - Gain in PE

yes

Energy lost = work done = change in Kinetic Energy

Change in kinetic energy = (1/2 x 3.5 x 12^2) - (1/2 x 3.5 x 8^2)
which gave me 140J

I don't see now how I can use this. I can work out the force of the particle going down the slope, which would be 140/14 (since 14 is the displacement, and using the formula work done = force x displacement)

Then this means that Fmax would be 10, since it's decelerating, meaning that friction is at it's maximum? Is it right doe? This is the part where I get stuck.

I would need the height for that. I haven't been given that,

You haven't considered the gain in gravitational potential energy:

Energy lost = Loss in KE - Gain in PE

energy lost has to be positive right?

My loss in KE was 140J, and my gain in PE, by using P.E = mgh, where h is sin30x14, i get gain in P.E as (49/2)g, which means that as 140 - (49/2)g would be energy lost, which comes out negative.

I dont think I get this topic very well. Sorry. However isn't the energy loss meant to be positive?

You know it's travelled 14m so you can use trig with the angle to find the change in height.

Have you practiced basic questions before trying this one? It sounds like you're missing some of the basics.

How did you get 140 as loss in KE? Can you please show working for this?

It is a confusing topic at first but I'm sure with practice you'll be fine.

ok for some reason the angle I put down was 30 degrees, and it was actually 20 degrees.

So from the actual angle,

total energy loss = K.E lost - P.E gained

which goes to ((1/2 x 3.5 x 12^2)-(1/2 x 3.5 x 8^2))-3.5 x 9.8 x 14sin20

this however gives me the answer of -24.2 (3 s.f), which is defintely wrong.

The original answer is 304, so i don't know what I'm doing wrong.

Sorry I made a mistake in my post earlier. The particle is travelling down the slope so it is losing GPE. So

Total energy lost = loss in KE + loss in PE

ok that cleared all my working outs and gave me the correct answer

since it's loss in KE + loss in PE, Total energy loss is 304J (3 s.f), and carrying that value over to the formula F = W.D x S, where F is Fmax, since it's moving, we get Fmax as 21.73....

This means that using trig, we can get R as 3.5gCos20, since R is the same as the weigh (as it's same position horizontally), so by using those 2 values, I get Mu as 0.674, which is correct.

Thanks!