A mass m, lying on a horizontal frictionless surface is connected to mass M as shown in figure. The system is now released. The velocity of mass m when mass M as descended a distance h is
Dear Student,
Please find below the solution to the asked query:
When the mass 'M' has descendeda distnace of 'h', then the force acting on it will be its own weight ansdits acceleration will be 'g'. then,
from the second equation of motion:
h = ut + 1/2 gt2
u = 0
h = 0 +1/2gt2
t = √(2h/g)
Now,
acceleration of the mass 'm' = a = Mg/m
since initially everything is at rest then um = 0 m/s
now,
from the first equatin of motion if vm is the velocity of mass 'm' after time 't' when the mass M has fallen height 'h', then
vm = um + at
vm = 0 + (Mg/m)×(√(2h/g)
vm = √(2ghM2/m2)
Hope this information will clear your doubts about (work power and energy).
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Satyendra Singh
Please find below the solution to the asked query:
When the mass 'M' has descendeda distnace of 'h', then the force acting on it will be its own weight ansdits acceleration will be 'g'. then,
from the second equation of motion:
h = ut + 1/2 gt2
u = 0
h = 0 +1/2gt2
t = √(2h/g)
Now,
acceleration of the mass 'm' = a = Mg/m
since initially everything is at rest then um = 0 m/s
now,
from the first equatin of motion if vm is the velocity of mass 'm' after time 't' when the mass M has fallen height 'h', then
vm = um + at
vm = 0 + (Mg/m)×(√(2h/g)
vm = √(2ghM2/m2)
Hope this information will clear your doubts about (work power and energy).
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Satyendra Singh