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- #1

a) work out the tension in the string.

After 12s the block of mass 4kg reaches the bottom of the slope. The other block has not yet reached the pulley.

b) work out a lower bound for the length of the string giving your ans to 2 sf

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- Thread starter
- #1

a) work out the tension in the string.

After 12s the block of mass 4kg reaches the bottom of the slope. The other block has not yet reached the pulley.

b) work out a lower bound for the length of the string giving your ans to 2 sf

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- #2

A block of mass 4kg is held on a rough slope that is inclined at 30 degree to the horizontal. The coefficient of friction between the slope and the block is 0.2. A light inextensible string is attached to the block and runs parallel to the slope to pass over a small smooth pulley fixed at the top of the slope. The other end of the string hangs vertically with a block of mass 1kg attached to the other end. The system is released from rest.

a) work out the tension in the string.

After 1.2s the block of mass 4kg reaches the bottom of the slope. The other block has not yet reached the pulley.

b) work out a lower bound for the length of the string giving your ans to 2 sf

I got the ans for (a) which is T= 10.6

I don't understand how to calculate (b)

Last edited:

- Mar 1, 2012

- 935

$L > \dfrac{1}{2}at^2$

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- #4

T= 10.6N, a=-0.6m/s^2, I get s=0.43$L > \dfrac{1}{2}at^2$

- Mar 1, 2012

- 935

s= 0.43 ?T= 10.6N, a=-0.6m/s^2, I get s=0.43

you calculated that value using t = 12 seconds, or was that a typo in your original post?

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- #6

I calculated using t=1.2ss= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?

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- #7

Iam sorry that was a typo error. t= 1.2 s not 12ss= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?

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- #8

Iam getting the ans of s=0.43 m but the ans in the textbook is 0.66ms= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?

- Mar 1, 2012

- 935

If that is also the length of the incline, then the height of the incline is 0.22 m.

The only way I can explain the “text” answer is that the two were added, yielding 0.66 m, which doesn’t make much sense to me. That would essentially say the small mass is still at the bottom of the vertical edge of the incline when the large mass is at the bottom of the slanted part of the incline … go figure

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- #10

I agree with you. Thank you so much.

If that is also the length of the incline, then the height of the incline is 0.22 m.

The only way I can explain the “text” answer is that the two were added, yielding 0.66 m, which doesn’t make much sense to me. That would essentially say the small mass is still at the bottom of the vertical edge of the incline when the large mass is at the bottom of the slanted part of the incline … go figure