A bullet from speeding to full stop.

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Homework Help Overview

The discussion revolves around a physics problem involving a bullet with a mass of 15 g moving at a speed of 425 m/s that comes to a stop in a block of wood over a distance of 3.5 cm. Participants are tasked with finding the force acting on the bullet, assuming it is constant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore which physical laws may apply to the situation, questioning the role of gravitational force in the context of work done. There is discussion about the equation to use, with some suggesting a focus on kinetic energy and force without considering gravitational work.

Discussion Status

There is an ongoing exploration of the implications of negative force values, with some participants clarifying that the negative sign indicates direction rather than an error in calculation. The conversation also touches on the importance of significant figures in reporting the final answer, with various interpretations being discussed.

Contextual Notes

Participants are navigating the complexities of significant figures as they relate to the problem's parameters, which are given with varying levels of precision. There is uncertainty regarding how to properly express the final force value based on these significant figures.

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Homework Statement


A bullet (mass 15 g) moving at a speed of 425 m/s is brought to a stop in 3.5 cm in a block of wood.

Homework Equations


Find the force on the bullet, assuming that it is constant.


The Attempt at a Solution


what law should we use?
0 - 1/2*m*V^2 = m*g*0.035 + F*0.035 ?
 
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chawki said:

Homework Statement


A bullet (mass 15 g) moving at a speed of 425 m/s is brought to a stop in 3.5 cm in a block of wood.

Homework Equations


Find the force on the bullet, assuming that it is constant.


The Attempt at a Solution


what law should we use?
0 - 1/2*m*V^2 = m*g*0.035 + F*0.035 ?
Explain why you think that the gravity force does work.
 
It's a law..isn't ? HELP
 
Wow. I think i was wrong...it doesn't have work because the motion is on X axis, not Y axis.
so i think we will have only: - 1/2*m*V^2 = F*0.035 ?
 
chawki said:
so i think we will have only: - 1/2*m*V^2 = F*0.035 ?

yup! :smile:
 
But then we will get F=-38705.35N
How is possible to have a negative value!
 
Force is a vector, that has both magnitude and direction. The negative sign implies that if the bullet is moving from left to right, in the positive x direction, then the force acting on it is in the opposite (negative x) direction, from right to left, acting to slow it down.
 
Ok, so F=-38705.35N is correct ?
And thank you everyone, I'm learning new things :)
 
chawki said:
Find the force on the bullet, assuming that it is constant.
chawki said:
Ok, so F=-38705.35N is correct ?

no, the question is either asking for the magnitude of F, which is always positive, or it's asking for a "vector" description of F, which means a magnitude and a direction

(oh, and far too many significant figures :redface:)
 
  • #10
ok, so it is 38705.35N!
 
  • #11
yeees …

but what did i say about the significant figures?
 
  • #12
38705?
 
  • #13
hmm …

how many significant figures are there in the question?

so how many do you think there should be in the answer? :smile:
 
  • #14
I never ever used these things of significant figures...that's why I'm all confused
 
  • #15
ok …*look at signifcant figures in the PF Library :smile:
 
  • #16
38705 Yes or NO
 
  • #17
tiny-tim said:
hmm …

how many significant figures are there in the question?

so how many do you think there should be in the answer? :smile:

There are three? so the answer would be 38705.357N?
 
  • #18
chawki said:
There are three?

no, the question gives m v and d to 3 and 2 significant figures …

so the answer is probably best in only 2 sig figs
… so the answer would be 38705.357N?

are you even trying? :redface:

how would you write 38705.357 to 3 sig figs?

how would you write 38705.357 to 2 sig figs?

(if you don't know how, then look it up)
 
  • #19
39000n?
 
  • #20
yes, to 3 significant figures it's 38700, to 2 it's 39000
 
  • #21
we can write also: - 1/2*m*V^2 = -F*0.035 knowing that F will be from right to left, opposite the bullet direction and then we will find F the same. no need to introduce the magnitude.
 

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