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M1 - Friction question

Having problems thinking how to do part 2 of this question:

https://gyazo.com/0882b963d749914320663811225e518e

Ty!
Original post by Jian17
Having problems thinking how to do part 2 of this question:

https://gyazo.com/0882b963d749914320663811225e518e

Ty!


Hello! Can you post your working for part 1 so it's easier to help you with the second part :smile:
Reply 2
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Jian17
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image-ce63051f-e9d1-4880-9f5e-49c0a2266aa8753081731-compressed.jpg.jpeg Working out
There are two errors in your working for part (i) that seem to have cancelled each other out to give the correct coefficient of friction. While the particle is travelling up the slope, the frictional force and the component of the particle's weight parallel with the slope both act DOWN the slope and will thus have the same sign. If we use down the slope as the positive direction, mg.sin(18) + Friction = 2.4 (where 2.4 is the net force acting on the particle in the direction of the slope, consistent with a deceleration of 4ms^-2). You have written -mg.sin(18) + Friction = 2.4 but still somehow came out with the right answer!

For part (ii), once the particle starts to move down the slope, the component of the particle's weight parallel with the slope will continue to act DOWN the slope but the frictional force will no act UP the slope (resisting the particle's motion) , so the two will now be opposite in sign.
Reply 4
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Jian17
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Could you draw the diagram for the question for me please?
Original post by Jian17
Could you draw the diagram for the question for me please?


The diagram you posted earlier looks pretty good for part (ii) of the question. You have a component of the particle's weight acting down the slope and friction acting up the slope, which is correct for motion down the slope. To complete the diagram for the purposes of part (ii), you could add the value of "mu" you found in part (i) and delete a=4 as the acceleration is unknown in part (ii).

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