two masses m1 and m2 are attached to a spring balance S.if m1 is greater than m2 then the reading of spring balance will be
Answers
Answered by
68
Given -
m1 > m2
Reading on spring balance = ?
Solution-
Since m1 > m2, there will be an acceleration in the system.
m1g-m2g = (m1+m2)a
a= (m1-m2)g/(m1+m2)
Restoring force,
R= T1 = T2
Thus, R = T1 = m1(g-a)
R = m1[g- (m1-m2)g/(m1+m2) ]
R = m1g (m1+m2-m1+m2)/(m1+m2)
R = 2m1m2g/(m1+m2)
The reading of the spring balance is nothing but restoring force.
Hence, reading on spring balance is 2m1m2g/(m1+m2).
Hope that was useful.
Answered by
5
Since m1 > m2, there is an acceleration in the system
- m1g-m2g = (m1+m2)a
- a= (m1-m2)g/(m1+m2)
Restoring force,
- R= T1 = T2
- R = T1 = m1(g-a)
- R = m1[g- (m1-m2)g/(m1+m2) ]
- R = m1g (m1+m2-m1+m2)/(m1+m2)
- R = 2m1m2g/(m1+m2)
I hope that this was useful for you.
Similar questions