You can think of the work-kinetic energy theorem as the second theory of motion, parallel to newton's laws in describing how outside influences affect the motion of an object. in this problem, solve parts (a), (b), and (c) separately from parts (d) and (e) so you can compare the predictions of the two theories. a 16.1 g bullet is accelerated from rest to a speed of 730 m/s in a rifle barrel of length 64.8 cm.

(a) find the kinetic energy of the bullet as it leaves the barrel.

(b) use the work-kinetic energy theorem to find the net work that is done on the bullet.

(c) use the result in part (b) to find the magnitude of the average net force that acted on the bullet while it was in the barrel.

(d) now model the bullet as a particle under constant acceleration. find the constant acceleration of the bullet that starts from rest and gains a speed of 730 m/s over a distance of 64.8 cm.

(e) modeling the bullet as a particle under a net force, find the net force that acted on it during its acceleration.

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

    m = 16.1 g = 1.61 x 10^-2 kg

     v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest)  Ki = 0, therefore:

W = ΔK = Kf - Ki  = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit