Calculating Average Net Force Exerted on a 15-g Bullet Fired from a Rifle

[courtney101ann]

A 15-g bullet is fired from a rifle. It takes 2.5 × 10^-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s . Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet
Answer : 4290 N
I need step by step explains please

 

[adjacent]

Welcome to PF :)
Please Show what you have done so far.

 

[klimatos]

Neither the acceleration nor the speed can remain constant. The acceleration ceases as soon as the propellant stops acting on the bullet. The bullet speed will diminish steadily due to friction with the surrounding air.

 

[adjacent]

Neither the acceleration nor the speed can remain constant. The acceleration ceases as soon as the propellant stops acting on the bullet. The bullet speed will diminish steadily due to friction with the surrounding air.

As this question does not mention friction,it should be neglected.

 

[collinsmark]

Hello courtney101ann,

Welcome to Physics Forums!

A 15-g bullet is fired from a rifle. It takes 2.5 × 10^-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s . Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet
Answer : 4290 N
I need step by step explains please

As adjacent points out, please show your work and maybe we can help you if you are stuck.

In the mean time, let me throw out a couple of questions for consideration:

• What is the change in the bullet's momentum?
• What is an impulse, and how is it defined? (Defined in the context of force, time, momentum, etc.)

[Edit: klimatos and adjacent: as it turns out, the acceleration doesn't need to constant (uniform) to determine the average, net force. The average, net force can be determined quite easily, in fact, even if the acceleration varies all over the place. The problem statement's assumption about the acceleration being constant is superfluous; it's an unnecessary assumption.

In addition, friction can be present too. It doesn't matter. Friction can be part of the "net" force. Take it or leave it. Whether friction is present or not doesn't change the final answer.]

 

[haruspex]

as it turns out, the acceleration doesn't need to constant (uniform) to determine the average, net force. The average, net force can be determined quite easily, in fact, even if the acceleration varies all over the place. The problem statement's assumption about the acceleration being constant is superfluous; it's an unnecessary assumption.

I presume you are suggesting using ΔE/Δs = ∫F.ds/∫ds. I'm afraid that would be an error. I see it several times a year in book questions posed on this forum.
Consider whether average force should be defined as ∫F.dt/∫dt or ∫F.ds/∫ds. I would argue that the appropriate definition would have to match that of average acceleration, namely Δv/Δt = ∫a.dt/∫dt. If acceleration is not constant then this will generally not be equal to ∫a.ds/∫ds.

 

[collinsmark]

I presume you are suggesting using ΔE/Δs = ∫F.ds/∫ds. I'm afraid that would be an error. I see it several times a year in book questions posed on this forum.
Consider whether average force should be defined as ∫F.dt/∫dt or ∫F.ds/∫ds. I would argue that the appropriate definition would have to match that of average acceleration, namely Δv/Δt = ∫a.dt/∫dt. If acceleration is not constant then this will generally not be equal to ∫a.ds/∫ds.

It's simpler than that. I'm just saying that the impulse equals the change in momentum.[Edit: but yes, if you wish, I'm also saying that the average, net force is $\vec F_{ave} = \frac{\int_{t_1}^{t_2} \vec F_{net}(t) \ dt}{t_2 - t_1}$ But that's a little more complicated than I wanted to get into. So I'll summarize that by saying the impulse is also equal to the average net force times the time interval in question; $\vec J = \vec p_2 - \vec p_1 = \Delta \vec p = \vec F_{ave} \ \Delta t$ ]

 

[haruspex]

It's simpler than that. I'm just saying that the impulse equals the change in momentum.


[Edit: but yes, if you wish, I'm also saying that the average, net force is $\vec F_{ave} = \frac{\int_{t_1}^{t_2} \vec F_{net}(t) \ dt}{t_2 - t_1}$ But that's a little more complicated than I wanted to get into. So I'll summarize that by saying the impulse is also equal to the average net force times the time interval in question; $\vec J = \vec p_2 - \vec p_1 = \Delta \vec p = \vec F_{ave} \ \Delta t$ ]

My mistake - I didn't read the information provided properly. I thought it gave the length of the barrel, not the time.

 

[Chestermiller]

A 15-g bullet is fired from a rifle. It takes 2.5 × 10^-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s . Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet
Answer : 4290 N
I need step by step explains please

What is the speed at time zero?
What is the speed at time 2.5 × 10^-3 s?
What is the acceleration?
What is the relationship between force, mass, and acceleration?
What is the force?

 

[adjacent]

It seems the OP is inactive.
However,only after answering @Chestermiller's questions,should any help be provided.

 

[courtney101ann]

I can't answer @Chestermiller's question I only have the information I posted and as for work I don't have any I'm completely clueless we only got half way through the notes when we had an emergency release due to the snow and we haven't returned to school yet and my teacher still wants it done by Monday

 

[courtney101ann]

I do have my free body diagram and summations

 

[courtney101ann]

This is the work

 

[Chestermiller]

I can't answer @Chestermiller's question I only have the information I posted and as for work I don't have any I'm completely clueless we only got half way through the notes when we had an emergency release due to the snow and we haven't returned to school yet and my teacher still wants it done by Monday

Do you have any idea what the answers to my first two questions are? The answers to these are mentioned right in your problem statement.

Chet

 

[adjacent]

@courtney101ann
Just sitting idle and saying "help me" is not a wise way.You have to think yourself.

P.S if you have completed Gr.8,You should be able to answer all of the Chestermiller's questions.