Average force on a 16g bullet accelerated to 1010m/s using work-energy theorem

[blayman5]

Homework Statement

A 16 g bullet is accelerated in a rifle barrel
72.3 cm long to a speed of 1010 m/s.
Use the work-energy theorem to find the
average force exerted on the bullet while it is
being accelerated. Answer in units of N.

Homework Equations

W=changeKE

The Attempt at a Solution

Fx=(mvf^2)/2-(mvi^2)/2
F=((16*1010^2)/2*.723
F=1.1287e7
Did I do this correctly?

 

[Stovebolt]

Homework Statement

A 16 g bullet is accelerated in a rifle barrel
72.3 cm long to a speed of 1010 m/s.
Use the work-energy theorem to find the
average force exerted on the bullet while it is
being accelerated. Answer in units of N.

Homework Equations

W=changeKE

The Attempt at a Solution

Fx=(mvf^2)/2-(mvi^2)/2
F=((16*1010^2)/2*.723
F=1.1287e7
Did I do this correctly?

Check your units. Looking at your equation

F=((16*1010^2)/2*.723

you have units of

g * m³ / s²


You seem to have the right equation for Kinetic Energy (although wrong units). That is, KE = (mass) * (velocity²) / 2, which should result in units of N * m.

 

[blayman5]

The way I had it in units was:

N * m = (kg *(m^2/s^2) )/ 2
N = (kg *(m^2/s^2) )/ 2 m
N = (kg *(m/s^2) )/ 2
N= kg * m/s^2

The method in units was correct but i don't know if there is an alternative method to solve this problem.

 

[borgwal]

almost correct: you should convert 16g to kg, that's all.

 

[blayman5]

ok, i got 11, 287.4 N thanks

 

[Axel92]

I have a problem that is quite similar.
But I am confused in the way you were able to find your answer.
when I multiplied your .016kg by 1010^2, I came up with 16321.6
which is not your 11,287.4 N

please explain.

 

[blayman5]

Remember that F=Wx, so find the work done by kinetic energy

N * m = (kg *(m^2/s^2) )/ 2
N = (kg *(m^2/s^2) )/ 2 m
N = (kg *(m/s^2) )/ 2

Use the distance given to find the force

 

[Axel92]

Yeah, shortly after I studdied the thread and got it all kinked out.
I love this place.
Thank you anyways.