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

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Homework Help Overview

The problem involves a bullet impacting a wooden block, where the goal is to determine the maximum force exerted on the bullet during its deceleration within the block. The context is rooted in concepts of momentum, work, and energy in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between momentum and work, with attempts to calculate the work done on the bullet and its kinetic energy. Questions arise about the correct units for momentum and the appropriate method for calculating the force.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some have raised concerns about the initial calculations and units used, while others are questioning how to effectively apply work and energy principles to find the maximum force.

Contextual Notes

There are indications of confusion regarding the units of momentum and the relationship between force, work, and energy. Participants are also grappling with the correct interpretation of the problem setup and the calculations involved.

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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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