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Two identical 5kg blocks are moving with same speed of 2m/s towards each other along a frictionless horizontal surface . The blocks collide , stick together and comes to rest . Considering the two blocks as a system , calculate the work done by (1) external forces (2) internal forces

Physics
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work done by all the forces= change in the kinetic energy
okay

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Other answers:

so what are the external forces exerting on these two blocks?
i did'n get the question , what external and what internal ?
gravity ?
there is no friction...so only external forces are gravity and normal reaction
okay
internal forces..when they'll collide with each other then they will exert forces on each other..that's the internal force
okay
it's fine?
ok
now will the external forces( gravity and normal reaction) do work?
no ?
|dw:1349445206541:dw|
yes...they wont do any work because they are perpendicular to d displacement...so no work done
okay understood
so from work energy theorem work done= change in the kinetic energy --> work done by external forces+ work done by the internal forces= change in the kinetic energy of the system
so u can calculate the work done by the internal forces
u know the work done by the external forces(=0) and the change in the KE of the system
okay
u'll calculate the change in the KE of the system?
final KE=0 initial KE= 1/2 m1 v1^2 + 1/2 m2 v2^2
okay
m1= m2=5kg and v1=v2=2m/s
20 J ?
yes
so work done by internal forces= (final KE- initial KE)= -20 J
okay
u got the meaning of -ve sign? I mean...why work done is -ve?
yes 0-20 ?
-ve work done because they r pulling down the energy of the system to zero
okay .. i understand ! thank you :)
:)

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