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- #1

- Thread starter Shah 72
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- #1

- Mar 1, 2012

- 935

You really need to figure out how to post a **readable** image ...

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

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- #3

I got the ans for this.You really need to figure out how to post areadableimage ...

View attachment 11169

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

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- #4

Iam still having doubts with q(b)You really need to figure out how to post areadableimage ...

View attachment 11169

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

For q(a)

I got a= 2m/s^2 and speed of p when q reaches the pulley = 1m/s

Q(b)Time when the system is released v= u+at1

t1=0.5s

After the string breaks, there is constant speed

I don't understand

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- #5

readableimage ...

View attachment 11169

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

- Mar 1, 2012

- 935

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$

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- #7

The ans is 0.1s but the text book says 0.9sand is in a state of free fall.

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$

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- #8

Oh I got it. It will be a quadratic equation and I solve using quadratic formulaand is in a state of free fall.

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$

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- #9

Thank you so much. t2= 0.4s so total time will be 0.9 sand is in a state of free fall.

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$