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anonymous
 4 years ago
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
anonymous
 4 years ago
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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0work done by all the forces= change in the kinetic energy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so what are the external forces exerting on these two blocks?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i did'n get the question , what external and what internal ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is no friction...so only external forces are gravity and normal reaction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0internal forces..when they'll collide with each other then they will exert forces on each other..that's the internal force

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now will the external forces( gravity and normal reaction) do work?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349445206541:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes...they wont do any work because they are perpendicular to d displacement...so no work done

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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
 4 years ago
Best ResponseYou've already chosen the best response.0so u can calculate the work done by the internal forces

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u know the work done by the external forces(=0) and the change in the KE of the system

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u'll calculate the change in the KE of the system?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0final KE=0 initial KE= 1/2 m1 v1^2 + 1/2 m2 v2^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0m1= m2=5kg and v1=v2=2m/s

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so work done by internal forces= (final KE initial KE)= 20 J

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u got the meaning of ve sign? I mean...why work done is ve?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ve work done because they r pulling down the energy of the system to zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay .. i understand ! thank you :)
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