What is the maximum force exerted on a bullet lodged in a block?

  • Thread starter Thread starter kminkel
  • Start date Start date
  • Tags Tags
    Block Bullet
Click For Summary

Homework Help Overview

The problem involves a bullet of mass 10.0 g traveling at 200 m/s that strikes a fixed wooden block and comes to rest 22 cm inside it. Participants are tasked with finding the maximum force exerted on the bullet by the block, as represented in a graph of force versus distance.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using work-energy principles, specifically relating work done to the kinetic energy of the bullet. There are attempts to apply the work equation W=Fd and to consider the integral of force over distance due to the variable nature of the force.

Discussion Status

Several participants provide guidance on using the integral to find the area under the force curve and equating it to the kinetic energy. There is an ongoing exploration of the correct bounds for integration and the interpretation of the graph. Some participants express confusion about the integration process and the calculations involved.

Contextual Notes

Participants are navigating the complexities of integrating a variable force and are clarifying the relationship between the area under the curve and the work done. There is mention of specific values and calculations that have led to confusion, indicating a need for careful interpretation of the problem setup.

kminkel
Messages
4
Reaction score
0

Homework Statement



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. Find the value of Fmax.

http://www.scribd.com/doc/135229669/Physics

Homework Equations



W=Fd
W=KE=.5mv^2

The Attempt at a Solution



I tried to treat is as a work problem and set Fd=.5mv^2.
I got an answer, but it wasn't one available to me.
I don't know what I am missing
 
Physics news on Phys.org
welcome to pf!

hi kminkel! welcome to pf! :smile:
kminkel said:
W=Fd

use the more general work done = integral of force "dot" distance :wink:

(W = ∫ F.ds)

show us what you get :smile:
 
Your idea is a good one, except the equation is only Fd for constant F. The graph shows F varies with distance, so what you're really doing is integral of F*dx.

Find the area under the curve (triangular + rectangular regions) and try setting that equal to the KE.

Edit: Looks like I was beaten to it!
 
With using the W=.5mv^2, I got the W=200 J.
But with the integral what would I use as the bounds? Would it be from 0 to .11 m or 0 to .22 m? Then how would you get rid of the integral symbol to solve for F. I'm sort of really confused.
 
You are integrating over the whole distance that the force is being applied (0 to 0.22m). If integrals are unfamiliar to you, just think of it like area.

Set the area under the curve from 0 to 0.22m equal to KE.
 
If I set the areas equal to .5mv^2...
.11x+.5(.22)x=.5(.01)(200)^2
.11x+.11x= 200
.22x=200
x=909.09
That isn't an option /:
 
kminkel said:
But with the integral what would I use as the bounds?

uhh? :confused: don't make it so complicated! :biggrin:

the integral of a curve is the area under it :smile:

(and yes, sometimes physics questions really are that simple! :wink:)
 
kminkel said:
If I set the areas equal to .5mv^2...
.11x+.5(.22)x=.5(.01)(200)^2
.11x+.11x= 200
.22x=200
x=909.09
That isn't an option /:


I see my error. I have the wrong base for the triangle. I fixed it and got 1212.12 which is an option! Thank you Both so much!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
7K
Replies
14
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
5
Views
4K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
1
Views
2K