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Question

An open knife edge of mass m is dropped from a height h on a wooden floor. If the blade penetrates up to the depth d into the wood, the average resistance offered by the wood to the knife edge is

A
mg
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B
mg(1hd)
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C
mg(1+hd)
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D
mg(1+hd)2
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Solution

The correct option is C mg(1+hd)

Initialy the knife have only potential energy i.e
U=mgh
The point when knife is just about to penetrate the wooden block it has only kinetic energy i.e.
K.E.=12mv2
Let it penetrates a distance d in the block then total K.E. will be zero and hence it will stop i.e vf=0
Using third law of motion:
v2u2=2aavgd
v=0
aavg=u22d --- (1)
Free body diagram of knife


Rmg=maavg

R=maavg+mg -- (2)

From eq.(1) and eq.(2)
R=m[g+u22d]
R=m[g+2gh2d]
R=mg[1+hd]

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