2 questions: Force of bullet and car

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    Bullet Car Force
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SUMMARY

The discussion addresses two physics problems: calculating the force exerted on a bullet and determining the resultant force and acceleration of a car. For the bullet, the force exerted while traveling down a 0.82m barrel at a speed of 320m/s is calculated to be 624N, using the formula F=ma with an acceleration of 124800m/s². The second problem involves two forces acting on a car with a mass of 3000kg, but the solution requires a free body diagram to determine the resultant force and acceleration accurately.

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Homework Statement


(1)A 5g bullet leaves the muzzle of a rifle with a speed of v=320m/s. what force (assumed constant) is exerted on the bullet while it is traveling down the 0.82m-long barrel of the rifle?

(2)Two forces are applied to a car in an effort to move it. what is the resultant of these two forces? if the car has a mass of 3000kg, what acceleration does it have? ignore friction.
The car is facing 90degree position, 1st force is 60degrees at 400N, 2nd force is at 100degrees at 450N.

Homework Equations


d=vt
a=v/t
f=ma


The Attempt at a Solution


(1) i tried, d=vt, .82=320m/s*t, t=.003s
a=v/t, a=320m/s / .003s, a=124800m/s^2
f=ma, F= .005kg*124800m/s^2, F=624N

(2) i don't know how to do this one..
 
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for 2) draw the free body diagram first(and post it here) and see if you get an idea.
 
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mikex213 said:

The Attempt at a Solution


(1) i tried, d=vt, .82=320m/s*t, t=.003s
The v in that formula should be vaverage. However, you have used vfinal.

What is vaverage, given that vinitial is zero and vfinal is 320 m/s/

a=v/t, a=320m/s / .003s, a=124800m/s^2
f=ma, F= .005kg*124800m/s^2, F=624N
The rest of your work is correct, you just need to get the correct value for t first.
 

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