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 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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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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