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Physics Connected particles

Shah 72

Member
Apr 14, 2021
193
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 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
 

Shah 72

Member
Apr 14, 2021
193
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:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
$L > \dfrac{1}{2}at^2$
 

Shah 72

Member
Apr 14, 2021
193

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
T= 10.6N, a=-0.6m/s^2, I get s=0.43
s= 0.43 ?

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

Shah 72

Member
Apr 14, 2021
193
s= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?
I calculated using t=1.2s
 

Shah 72

Member
Apr 14, 2021
193
s= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?
Iam sorry that was a typo error. t= 1.2 s not 12s
 

Shah 72

Member
Apr 14, 2021
193
s= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?
Iam getting the ans of s=0.43 m but the ans in the textbook is 0.66m
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
Per my calculations, the 4kg mass moves a distance of 0.44 m.

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 :rolleyes:
 

Shah 72

Member
Apr 14, 2021
193
Per my calculations, the 4kg mass moves a distance of 0.44 m.

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 :rolleyes:
I agree with you. Thank you so much.