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- Mar 1, 2012

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same concepts apply to this problem as with the other posted pulley problem ...

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Using s= 1/2at^2, 5=1/2×8×t^2same concepts apply to this problem as with the other posted pulley problem ...

t=1.12s

After this how do I calculate? It's confusing.

- Mar 1, 2012

- 935

$T - f_k = ma_1$

$Mg - T = Ma_1$

The second acceleration, $a_2$, is only for the smaller mass …

$-f_k = ma_2$

The small mass moves only 1m with acceleration $a_1$. After moving that 1m, tension becomes zero when the larger mass hits the ground. The smaller mass continues moving with acceleration $a_2$ until it comes to a stop.

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- #5

$T - f_k = ma_1$

$Mg - T = Ma_1$

The second acceleration, $a_2$, is only for the smaller mass …

$-f_k = ma_2$

The small mass moves only 1m with acceleration $a_1$. After moving that 1m, tension becomes zero when the larger mass hits the ground. The smaller mass continues moving with

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- #6

So a= 8m/s^2

$T - f_k = ma_1$

$Mg - T = Ma_1$

The second acceleration, $a_2$, is only for the smaller mass …

$-f_k = ma_2$

The small mass moves only 1m with acceleration $a_1$. After moving that 1m, tension becomes zero when the larger mass hits the ground. The smaller mass continues moving with acceleration $a_2$ until it comes to a stop.

S= ut +1/2at^2

1=1/2×8×t^2, t=0.5s

I calculated v=4m/s

F=m×a

-5=0.5×a

a=-10m/s^2

I still don't get the ans mentioned in the textbook which is t= 2.5s

- Mar 1, 2012

- 935

$a_1= 8 \, m/s^2 \implies t_1 = 0.5 \, s \implies v_f = a_1 t_1 = 4 \, m/s$

$v_f = 4 \, m/s$ becomes $v_0$ for the second part …

$a_2 = -\mu g = -2 \, m/s^2$

$v_f = v_0 + a_2t_2 \implies 0 = 4 - 2t_2 \implies t_2 = 2 \, s$

$t_1+t_2 = 2.5 \, s$

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Thank you so so so so so so much!!!!!

$a_1= 8 \, m/s^2 \implies t_1 = 0.5 \, s \implies v_f = a_1 t_1 = 4 \, m/s$

$v_f = 4 \, m/s$ becomes $v_0$ for the second part …

$a_2 = -\mu g = -2 \, m/s^2$

$v_f = v_0 + a_2t_2 \implies 0 = 4 - 2t_2 \implies t_2 = 2 \, s$

$t_1+t_2 = 2.5 \, s$