A train is travelling up a 3.73 degree incline at a speed of 3.25 metres per second, when the last cart breaks free and begins to coast without friction.
(I) How long does it take for the last cart to come to rest momentarily (ii) How far did the last cart travel before (momentarily) coming to rest (iii) For a real train, the friction between the cart and the track can be described by a friction coefficient. Find this coefficient, assuming that the time it takes for the cart to come to rest is 3 seconds.

Post your attempt at the question please.

Hint: Use energy instead of kinematics

Ok, thanks - not sure how to apply energy equations here though. The way we were taught is that for part (I), using Newton's Laws: mg sin theta = ma. (Equation 1). So a = g sin theta = 0.64 ms-2. I presume this is negative for deceleration (if the cart is continuing upwards for a short while, or moving against g).Then, using V = V(0) + at, where V = 0 and V(0) = 3.25 ms-1, I got 5.08 seconds for t. For part (ii), I used the other equation of motion, V squared = V(0) squared + 2ax to work out x, which gave me 8.25 metres. For part (iii), I firstly worked out what the acceleration would be if t was 3 seconds (and not 5.08 seconds as above). Then I added the extra term for the frictional force (Fk) to Equation 1 above: mg sin theta - Fk = ma. (perhaps this should be mg sin theta + Fk = ma (since both arrows for the first 2 terms would be pointing down the slope). However, if mg sin theta is negative (car is moving upwards against gravity), perhaps it should be - mg sin theta + Fk = ma?

Okay, I was in a hurry earlier and for the first one, you do have to use kinematics :o It's the second part that can be done independently from the first part using energy.

(i) Using g = 9.81 m/s^2, I get 5.09 s. I guess you used 9.8 m/s^2 as g. My comment on that part is that if you know g as 9.81 m/s^2 and have a question where the figures are given to 3 significant figures, you use 9.81 m/s^2. Otherwise, if the figures are given to 2 s.f. you can go safely with 9.8 m/s^2

(ii) Well, of course since I used 9.81 m/s^2, I got slightly larger value of 8.275.. = 8.28 m

Here's what you could do too if the question doesn't specific what method to use:

\frac12 mv^2 = mgh

(Kinetic energy lost by the cart is converted to gravitational potential energy and when there is no more kinetic energy, it means all the kinetic energy was converted to potential energy)
From there, the vertical height the cart travelled can be obtained by solving for h.

h = \frac{v^2}{2g}

Then, the distance along the slope is obtained by using trigonometry:

d = \frac{h}{\sin\theta} = \frac{v^2}{2g\sin\theta} = \frac{(3.25)^2}{2(9.81)\sin(3.73)} = 8.28\ m

(iii) Okay for this part, you started right.

Friction is against motion and so, it is also in the direction: down the slope so that you have Friction + Gravity Component = Net Force

All being in the same direction. Can you try that?

OK, thanks. Yes, using the energy equation for part (ii) makes sense. For part (iii), I guess it also makes sense that the friction force and the component of the force from gravity are added together since both arrows are pointing down the slope. I got confused with the cart moving upwards against gravity. So I got: mg sin theta + U(k)N = ma , where U(k) is the Coefficient of Kinetic Friction. This then gave me: mg sin theta + U(k) mg cos theta = ma. Cancelling out the masses, I got g sin theta + U(k) g cos theta = a (although I'm not sure if a is supposed to be negative if the cart is decelerating). Using the value of a = 1.083 seconds when t = 3 seconds (I worked this out from V = V(o) + at, where the final velocity V = 0 and the initial velocity V(o) = 3.25), I got 0.638 + 9.789 U(k) = 1.083, which gave me a value of 0.045 for the coefficient of kinetic friction (hopefully, I made no errors here)

I see no errors there. Well done! :)

That's great. Many thanks.