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ashna
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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
 2 years ago
 2 years ago
ashna Group Title
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
 2 years ago
 2 years ago

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akash123 Group TitleBest ResponseYou've already chosen the best response.1
work done by all the forces= change in the kinetic energy
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
so what are the external forces exerting on these two blocks?
 2 years ago

ashna Group TitleBest ResponseYou've already chosen the best response.0
i did'n get the question , what external and what internal ?
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
there is no friction...so only external forces are gravity and normal reaction
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
internal forces..when they'll collide with each other then they will exert forces on each other..that's the internal force
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
it's fine?
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
now will the external forces( gravity and normal reaction) do work?
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
dw:1349445206541:dw
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
yes...they wont do any work because they are perpendicular to d displacement...so no work done
 2 years ago

ashna Group TitleBest ResponseYou've already chosen the best response.0
okay understood
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
so u can calculate the work done by the internal forces
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
u know the work done by the external forces(=0) and the change in the KE of the system
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
u'll calculate the change in the KE of the system?
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
final KE=0 initial KE= 1/2 m1 v1^2 + 1/2 m2 v2^2
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
m1= m2=5kg and v1=v2=2m/s
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
so work done by internal forces= (final KE initial KE)= 20 J
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
u got the meaning of ve sign? I mean...why work done is ve?
 2 years ago

akash123 Group TitleBest ResponseYou've already chosen the best response.1
ve work done because they r pulling down the energy of the system to zero
 2 years ago

ashna Group TitleBest ResponseYou've already chosen the best response.0
okay .. i understand ! thank you :)
 2 years ago
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