A horizontal massive platform is moving with a constant velocity v at times t=0 a small block of mass m is gentky place on platform
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Answered by
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Because in ground frame the block of mass m is placed at platform with zero velocity, but finally it got some velocity.
That velocity must be same as the velocity of the platform because the friction force will stop acting when there is no relative motion.
Now only force doing work on the block, is friction.
By work energy theorem, change in kinetic energy is equal to the net work done.
Hence work done by friction is 1/2 mv^2 -0
ANSWER:
✔️+(1/2)mv^2 b) -(1/2)mv^2
That velocity must be same as the velocity of the platform because the friction force will stop acting when there is no relative motion.
Now only force doing work on the block, is friction.
By work energy theorem, change in kinetic energy is equal to the net work done.
Hence work done by friction is 1/2 mv^2 -0
ANSWER:
✔️+(1/2)mv^2 b) -(1/2)mv^2
Answered by
3
Explanation:
Because in ground frame the block of mass m is placed at platform with zero velocity, but finally it got some velocity.
That velocity must be same as the velocity of the platform because the friction force will stop acting when there is no relative motion.
Now only force doing work on the block, is friction.
By work energy theorem, change in kinetic energy is equal to the net work done.
Hence work done by friction is 1/2 mv^2 -0
ANSWER:
✔️+(1/2)mv^2 b) -(1/2)mv^2
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