The discussion revolves around calculating the average net force exerted on a 15-g bullet fired from a rifle, given the time it takes for the bullet to travel the barrel and its exit speed. The problem involves concepts from kinematics and dynamics, particularly relating to acceleration and force.
There is an ongoing exploration of the problem with various participants questioning the assumptions made in the original post. Some participants have offered guidance by suggesting relevant questions to consider, while others emphasize the need for the original poster to clarify their understanding and provide their work.
Some participants note that the original poster may lack sufficient information or understanding due to recent disruptions in their education, which could affect their ability to engage with the problem fully.
As this question does not mention friction,it should be neglected.klimatos said:
courtney101ann said:
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.collinsmark said:
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 said:
My mistake - I didn't read the information provided properly. I thought it gave the length of the barrel, not the time.collinsmark said:
What is the speed at time zero?courtney101ann said:
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.courtney101ann said:
