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# m1 equilibrium question

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1. I've managed to solve parts a and b but i do not understand how to do part c, i got a as t=1.2mg and b was sqrt(1.4gh). can someone explain to me how to do this?
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3. (Original post by jamesw5555)
I've managed to solve parts a and b but i do not understand how to do part c, i got a as t=1.2mg and b was sqrt(1.4gh). can someone explain to me how to do this?
Outline:

1) Q falls a distance of h so P moves a distance of h as well.

2) P's speed after moving h metres is sqrt(1.4gh) since they are connected.

3) The string now goes slack. So resolve to the right and use the fact that the only acceleration is now due to friction (it should be negative).

4) Use suvat with the acceleration in 3), the initial speed in 2) and a final velocity of 0 (at rest) to find the distance moved by P.

5) Since P moved a distance of h + answer to 4) and it hasn't reached the pulley, then...?
4. (Original post by Zacken)
Outline:

1) Q falls a distance of h so P moves a distance of h as well.

2) P's speed after moving h metres is sqrt(1.4gh) since they are connected.

3) The string now goes slack. So resolve to the right and use the fact that the only acceleration is now due to friction (it should be negative).

4) Use suvat with the acceleration in 3), the initial speed in 2) and a final velocity of 0 (at rest) to find the distance moved by P.

5) Since P moved a distance of h + answer to 4) and it hasn't reached the pulley, then...?
Thanks for the help I understood it perfectly, but the initial was 0 and final was 1.4gh as i got -1.4h when they were reversed.
5. (Original post by jamesw5555)
Thanks for the help I understood it perfectly, but the initial was 0 and final was 1.4gh as i got -1.4h when they were reversed.
Huh? No, the final is 0 and the initial is sqrt(1.4gh).

Then: v^2 = u^2 + 2as to get 0 = 1.4gh + 2as which then gets -2as = 1.4gh which then gets s = -1.4gh/(2a).

Remember that your a should be negative.
6. (Original post by Zacken)
Huh? No, the final is 0 and the initial is sqrt(1.4gh).

Then: v^2 = u^2 + 2as to get 0 = 1.4gh + 2as which then gets -2as = 1.4gh which then gets s = -1.4gh/(2a).

Remember that your a should be negative.
Oh never mind, thanks for clearing that up i looked up the mark scheme and it said 0^2 = 1.4gh - 2*0.5*h which i thought they had used v^2 - 2as =u^2
7. No problem.

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