anonymous
  • anonymous
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
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
work done by all the forces= change in the kinetic energy
anonymous
  • anonymous
okay

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

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