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Physics Mechanics- connected particles

Shah 72

Member
Apr 14, 2021
193
20210530_220932.jpg
I calculated a=8m/s^2. I don't understand how to calculate the total time.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
same concepts apply to this problem as with the other posted pulley problem ...
 

Shah 72

Member
Apr 14, 2021
193
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

Well-known member
MHB Math Helper
Mar 1, 2012
935
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

Member
Apr 14, 2021
193
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

Member
Apr 14, 2021
193
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

Well-known member
MHB Math Helper
Mar 1, 2012
935
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

Member
Apr 14, 2021
193
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!!!!!