Three identical blocks A, B, and C of masses m=2 kg each are connected by a light string sequentially where A and C are placed at the ends. A force of 10.2N is applied to pull at the end of C on a frictionless plane. What is the tension in the string between the blocks B and C?
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Answered by
11
hope this will help u
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Lipimishra2:
I am pretty sure what i typed from the paper is correct
...i dont want to point..but question should be wrong...beacuse ...it is givwn it is on friction less surface..so acceleration should be there..or it should be 0...or if friction less surfac is not there then we can take 9.8
I guess we need to take acc. 9.8, perhaps
....your teacger can also. do mistake..whatever..
Idk.
if you find the ans then inform me
Sure
Some were sayung it should be 6.8 or smthing.
Saying*
if a is 6.8 ..then gentelly put the acceleration value in the given equation..that i described
Answered by
19
see the attachment,
for block C,
forward force - backward force = net force
F - T2 = ma { let acceleration of system is a }
10.2 - T2 = 2a ----------------(1)
for block B,
forward -backward = ma
T2 - T1 = ma
T2 - T1 =2a-----------------------(2)
for block A,
similarly ,
T1 = ma
T1 = 2a ---------------(3)
now, from eqns (1) , (2) and (3)
10.2 = 6a
a = 10.2/6 = 1.7 m/s^2
hence, tension in string b/w the blocks B and C = T2 = 4a
= 4*1.7 = 6.8 N
for block C,
forward force - backward force = net force
F - T2 = ma { let acceleration of system is a }
10.2 - T2 = 2a ----------------(1)
for block B,
forward -backward = ma
T2 - T1 = ma
T2 - T1 =2a-----------------------(2)
for block A,
similarly ,
T1 = ma
T1 = 2a ---------------(3)
now, from eqns (1) , (2) and (3)
10.2 = 6a
a = 10.2/6 = 1.7 m/s^2
hence, tension in string b/w the blocks B and C = T2 = 4a
= 4*1.7 = 6.8 N
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thnx...dude
Thanks abhi!!!!
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