Resistance of wood to falling knife Watch

RyanCWY
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#1
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An open knife edge of mass M is dropped from a height h on a wooden floor. If the blade penetrates a distance s into the wood, the average resistance offered by wood to blade is?

The answer is Mg[1+(h/s)] but I don't get how.

I thought the equation should be
Mg(h+s) = KE right before hitting the wood + (Resistance x s)
and the KE right before hitting the wood would be equal to lost in potential energy for the height fallen, so KE = Mgh. Therefore, the equation becomes
Mg(h+s) = Mgh + (Resistance x s) so
[(Mgs)/s)] = Resistance

Could someone please tell me why I'm wrong?
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Physics Enemy
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(Original post by RyanCWY)
...
Work done bringing knife to rest (KE = 0) at x = 0 (GPE = 0), equals initial GPE of mg(h + s). So Fs = mg(h + s) => F = mg(1 + h/s).
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RogerOxon
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(Original post by RyanCWY)
Mg(h+s) = KE right before hitting the wood + (Resistance x s)
and the KE right before hitting the wood would be equal to lost in potential energy for the height fallen, so KE = Mgh. Therefore, the equation becomes
Mg(h+s) = Mgh + (Resistance x s) so
[(Mgs)/s)] = Resistance

Could someone please tell me why I'm wrong?
The knife loses Mg(h+s) of GPE, with no nett change in KE (it starts and finishes with none). The only other (other than by gravity) work done on it is by the resistance of the wood, Fs.
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RogerOxon
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(Original post by RyanCWY)
I thought the equation should be
Mg(h+s) = KE right before hitting the wood + (Resistance x s)
That's the error. The knife starts and ends with a KE of 0.

You cannot have the KE when the knife hits the wood in the equation because it came from the GPE on the LHS and is entirely lost by the end. Your equation should be:

Energy lost = negative work done

Note that the energy lost is entirely GPE, as the knife starts and ends with no KE.

The work done by the wood's resistance on the knife is negative, as the force is opposite to the direction of travel of the knife (it slows it). We account for the work being negative by putting the loss of GPE in the equation.
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niteninja1
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Surely it depends on the heat of the knife and it's material as well?
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RogerOxon
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(Original post by niteninja1)
Surely it depends on the heat of the knife and it's material as well?
We're given the depth to which the knife penetrates, which will vary with different materials / conditions.
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RyanCWY
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(Original post by Physics Enemy)
Work done bringing knife to rest (KE = 0) at x = 0 (GPE = 0), equals initial GPE of mg(h + s). So Fs = mg(h + s) => F = mg(1 + h/s).
I get it now thankss
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RyanCWY
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(Original post by RogerOxon)
That's the error. The knife starts and ends with a KE of 0.

You cannot have the KE when the knife hits the wood in the equation because it came from the GPE on the LHS and is entirely lost by the end. Your equation should be:

Energy lost = negative work done

Note that the energy lost is entirely GPE, as the knife starts and ends with no KE.

The work done by the wood's resistance on the knife is negative, as the force is opposite to the direction of travel of the knife (it slows it). We account for the work being negative by putting the loss of GPE in the equation.
Thankss
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Physics Enemy
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(Original post by RyanCWY)
KE right before hitting the wood ...
To add, if you want to do the Q by considering knife's point of contact with floor:

W (KE = GPE = 0) = KE (poc) + GPE (poc)
=> Fs = mgh + mgs = mg(h + s)
=> F = mg(1 + h/s)

I think this is what you were shooting for.
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