Question

For the karate chop in the figure (Figure 1) , assume that the hand has a mass of 0.40kg and that the speeds of the hand just before and just after hitting the board are 15m/s and 0, respectively.
A) What is the average force exerted by the fist on the board if the fist follows through, so the contact time is 2.8ms ?
B) What is the average force exerted by the fist on the board if the fist stops abruptly, so the contact time is only 0.26ms ?

Answer 1

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Answer 2

A) Force = rate of chane in momentum. Momentum=mv

Answer 3

Getting it ?

Answer 4

I tried that and it wasn't correct.

Answer 5

A was correct when i converted it from ms to s but b was not. It does not give me an answer.

Answer 6

Change in momentum = 15*0.4 = 6
a) Since the contact time is 2.8ms = 2.8*10\(^{-3}\) s. Thus rate of change of momentum is 6/2.8*10\(^{-3}\) = 2142 N

Answer 7

b) Here the hand stops abrupty and the time is 0.26ms = 0.26*10\(^{-3}\)s
Rate of change in momentum = 6/0.26*10\(^{-3}\) = 23076 N

Answer 8

I think this should be correct

Answer 9

Got it I missed a zero. Thanks!!

Answer 10

I was correct then ?