ashna
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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ashna
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@akash123
akash123
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work done by all the forces= change in the kinetic energy
ashna
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okay
akash123
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so what are the external forces exerting on these two blocks?
ashna
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i did'n get the question , what external and what internal ?
ashna
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gravity ?
akash123
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there is no friction...so only external forces are gravity and normal reaction
ashna
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okay
akash123
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internal forces..when they'll collide with each other then they will exert forces on each other..that's the internal force
ashna
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okay
akash123
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it's fine?
akash123
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ok
akash123
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now will the external forces( gravity and normal reaction) do work?
ashna
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no ?
akash123
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|dw:1349445206541:dw|
akash123
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yes...they wont do any work because they are perpendicular to d displacement...so no work done
ashna
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okay understood
akash123
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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
akash123
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so u can calculate the work done by the internal forces
akash123
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u know the work done by the external forces(=0) and the change in the KE of the system
ashna
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okay
akash123
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u'll calculate the change in the KE of the system?
akash123
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final KE=0
initial KE= 1/2 m1 v1^2 + 1/2 m2 v2^2
ashna
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okay
akash123
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m1= m2=5kg
and v1=v2=2m/s
ashna
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20 J ?
akash123
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yes
akash123
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so work done by internal forces= (final KE- initial KE)= -20 J
ashna
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okay
akash123
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u got the meaning of -ve sign? I mean...why work done is -ve?
ashna
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yes 0-20 ?
akash123
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-ve work done because they r pulling down the energy of the system to zero
ashna
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okay .. i understand !
thank you :)
akash123
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:)