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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
 one year ago
 one year 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
 one year ago
 one year 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
 one year ago

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

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

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

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

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

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

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

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

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

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

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