MHB Mechanics- connected particles

[Shah 72]

I calculated a=8m/s^2. I don't understand how to calculate the total time.

 

[skeeter]

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

 

[Shah 72]

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

Using s= 1/2at^2, 5=1/2×8×t^2
t=1.12s
After this how do I calculate? It's confusing.

 

[skeeter]

The problem requires calculation of two accelerations. The first, $a_1$, is determined by the equations

$T - f_k = ma_1$
$Mg - T = Ma_1$

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.

 

[Shah 72]

The problem requires calculation of two accelerations. The first, $a_1$, is determined by the equations

$T - f_k = ma_1$
$Mg - T = Ma_1$

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

 

[Shah 72]

The problem requires calculation of two accelerations. The first, $a_1$, is determined by the equations

$T - f_k = ma_1$
$Mg - T = Ma_1$

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.

So a= 8m/s^2
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

 

[skeeter]

correct on the first part …

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

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

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

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

$t_1+t_2 = 2.5 \, s$

 

[Shah 72]

correct on the first part …

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

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

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

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

$t_1+t_2 = 2.5 \, s$

Thank you so so so so so so much!