Mechanics A level question

Hi, can anyone do this question?

A particle of mass 4kg, initially at rest, is pulled up a rough surface, which is inclined at 25 degrees to the horizontal, by a force of 50N that acts at an angle of 30 degrees to the surface. coefficient of friction between particle and surface is 0.1

acceleration of particle I got as 6.42 ms^-2 which is right
2nd part of question is: after 3 seconds the force is removed, work out the time for the particle to come to rest. The answer is 5.92 seconds, but I'm not sure how to get the answer. I used suvat to work out speed before the force is removed, than worked out the acceleration without the 50N force up the slope, and substituted into another suvat. The answer I got is 3.83 s, can anybody help me with this question?

Haven't checked the first part, but using your value for the inital acceleration, I get the same time as you for the second part. Either they have the wrong answer, or we are both making the same mistake! Where is the question from?

I got the question from the pearson edexcel practice book for year 2 mechanics, it might be wrong, they have gotten answers wrong before , thanks for going through it though . Did you use the same method as I did or did you do it another way?

no the answer is correct. I wrote out the whole method but TSR striked me because i gave the answers as well :///

Basically you have to resolve to find the normal reaction force because when you first resolved it, the vertical component of 50N affected your result. Also because the 50N force has been removed you have to remember that friction will act in the opposite direction as the mass will start to slide down the slope.
Using your value for acceleration, plug it into the formula a=(v-u)/t and rearrange for v - this tells you the final speed of the mass before the 50N is removed.
Now using your new value for normal reaction, resolve and plug this into f=ma to calculate the new acceleration of the mass due to gravity (as the 50N is no longer acting on it)
use a=(v-u)/t again but remember that the value you just calculate as v now becomes u as the mass will eventually come to a stop meaning v=0
you should get 5.92s

Thank you! I was stuck on this question for a while, I understand now I'll try to remember that the friction always starts to oppose the direction of motion

From what I can see, your original answer for when the mass first comes to rest was correct, and the book is wrong.

@Lenaa_x 's idea that friction reverses when the force is removed, whilst it will give you the answer the book provides, is incorrect. The mass will be slowing down, but it will still be moving up the plane and friction will still be acting down the plane, until it comes to instantaneous rest.

Can you provide a link/image of the original question so we can see it exactly as written?

From what I can see, your original answer for when the mass first comes to rest was correct, and the book is wrong.

@Lenaa_x 's idea that friction reverses when the force is removed, whilst it will give you the answer the book provides, is incorrect. The mass will be slowing down, but it will still be moving up the plane and friction will still be acting down the plane, until it comes to instantaneous rest.

Can you provide a link/image of the original question so we can see it exactly as written?

But I think it starts moving downwards doesn't it? The only upwards force was the 50N force which is then removed

Just because the only forces acting are down the slope (both the component of weight down the slope and the friction), that doesn't mean that it will instantaneously start moving down the slope, only that it will accelerate down the slope. It is moving up the slope at the instant the 50 N force is removed, so it will have to slow down and stop before (possibly) moving back down the slope.

A similar but easier to understand situation is throwing a ball straight up. Once it has left your hand, there is no force pushing it up any more, only its weight pulling it down. There are no upwards forces, only downwards ones. Will the ball immediately start to move downwards? That would mean it is impossible to throw anything upwards!

Just because the only forces acting are down the slope (both the component of weight down the slope and the friction), that doesn't mean that it will instantaneously start moving down the slope, only that it will accelerate down the slope. It is moving up the slope at the instant the 50 N force is removed, so it will have to slow down and stop before (possibly) moving back down the slope.

A similar but easier to understand situation is throwing a ball straight up. Once it has left your hand, there is no force pushing it up any more, only its weight pulling it down. There are no upwards forces, only downwards ones. Will the ball immediately start to move downwards? That would mean it is impossible to throw anything upwards!

Ah I get the idea, but I guess in this A level question that idea might not be used? I don't know, I guess that because the course requires you to assume that friction is always in the opposite direction, you would use that in this question as well

But that is exactly what we are doing! In the first part of the question, when the particle was being pulled up the slope, you must have taken friction as acting down the slope, as it opposes the motion up the slope. So in the second part of the question, we take the friction down the slope as it opposes the motion of the particle which is still up the slope. It might not be being pulled up the slope any more, but it is still moving up the slope (well, until it comes to rest).

But that is exactly what we are doing! In the first part of the question, when the particle was being pulled up the slope, you must have taken friction as acting down the slope, as it opposes the motion up the slope. So in the second part of the question, we take the friction down the slope as it opposes the motion of the particle which is still up the slope. It might not be being pulled up the slope any more, but it is still moving up the slope (well, until it comes to rest).

From what I can see, your original answer for when the mass first comes to rest was correct, and the book is wrong.

@Lenaa_x 's idea that friction reverses when the force is removed, whilst it will give you the answer the book provides, is incorrect. The mass will be slowing down, but it will still be moving up the plane and friction will still be acting down the plane, until it comes to instantaneous rest.

Can you provide a link/image of the original question so we can see it exactly as written?

If you calculate the final velocity of the mass when the 50N force is still there, surely you can just plug it into suvat and reverse the motion of friction as the block will start to move down the slope bc of gravity?
How can the friction still work downwards when the 50N force is removed bc friction always opposes motion and without the 50N force, eventually the mass will start to move down the slope meaning friction must be working against it, no?

The key point is what you say at the end - eventually the mass will start to move down the slope. When that happens, then friction will oppose the motion and will act up the slope. But immediately after the force is removed, it is moving up the slope, and so friction will oppose this motion (and therefore act down the slope) until the mass comes to rest.

I agree that immediately after the force is removed the mass will continue moving up the slope but it will then stop and begin to slide down the slope. Can't you just switch the direction friction is working in once the mass initially comes to rest due the force being removed?

Yes, absolutely. Once the particle comes to rest, it will want to move down the slope so friction will act up the slope. But the part of the question we have been talking about concerns the motion from the point where the 50 N force is removed until the point when the particle comes to rest, and during this motion, it is still moving up the slope. Hence, we need to remember that friction still acts down the slope during this motion.

ah yes i see what you mean
so are you suggesting a new acceleration needs to be calculated after the 50N force has been removed and the mass comes to its first stop?