Fmax=36.4NWhat is the maximum force exerted on a bullet by a wooden block?

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A 10.0-g bullet traveling at 200 m/s strikes a wooden block and comes to rest 22 cm inside it. The maximum force exerted on the bullet by the block is being calculated, with initial discussions focusing on momentum and kinetic energy. The work done on the bullet, represented as the integral of force over distance, should equal the bullet's kinetic energy. There is confusion regarding the units of momentum and energy, with clarifications needed on the correct values. The conversation emphasizes the importance of using work and energy principles to find the maximum force exerted.
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A 10.0-g bullet traveling at 200 m/s strikes a fixed wooden block. The bullet comes to rest 22 cm inside the block. The magnitude of the force exerted on the bullet by the block over its 22-cm travel is shown in graph below. Find the value of Fmax.


my attempt:

p=mv=(0.01kg)(200m/s)=2J

(.22m)/(200m/s)=0.0011s

(0.0011/2)(fmax)+(.5)(0.0011/2)(fmax)=2J
----rectangle----------triangle--------total area----
 

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Momentum is not measured in Joules.
 
oops i guess it is Newton then but besides that am i right?
 
No, it is not Newton either. Momentum doesn't have named unit.

Given force and distance it will be easier to calculate work done, not change in momentum.
 
may i ask how would you start it with work and energy?
i know you can get the kinetic energy but what is next?
 
xstetsonx said:
may i ask how would you start it with work and energy?
i know you can get the kinetic energy but what is next?

The integral of the force over the distance is the work done on the bullet. As you seem to be computing. Shouldn't that equal the kinetic energy of the bullet? Which is not 2J.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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