you can think of the work-energy theorem as a second theory of motion, parallel

Question

you can think of the work-energy theorem as a 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 15.0 g bullet is accelerated from rest to a speed of 780 m/s in a rifle barrel of length 72.0 cm . (a) Find the kinetic energy of the bullet as it leaves the barrel. (b) Use the work-energy theorem to find the net work that is done on the bullet. ( c) Use your results from part (b) to find 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 a bullet that starts from rest and gains a speed of 780 m/s over a distance of 72.0 cm . (e) Use Newton’s second law to find the net force that acted on it during its acceleration. (f) What conclusion can you draw from comparing your results from parts (c) and (e)?

Explanation / Answer

here,

mass of the bullet , mb = 0.015 kg

speed of bullet , v = 780 m/s

length of barrel , l = 0.72 m

(a)

the kinetic energy of bullet , KE = 0.5 * m * v^2

KE = 0.5 * 0.015 * 780^2

KE = 4563 J

the kinetic energy of the bullet is 4563 J

(b)

using work energy theorm

work done = change in kinetic energy

work done = 4563 J

the net work done on the bullet is 4563 J

(c)

let the average force exerted be F

F * l*cos(0) = work done

F*0.72 = 4563

F = 6337.5 N

the average force exerted on the bullet is 6337.5 N

(d)

let the accelration be a

using third equation of motion

v^2 - u^2 = 2*a*l

780^2 - 0 = 2 * a*0.72

a = 422500 m/s^2

the accelration of the bullet is 422500 m/s^2

(e)

using Newton's seccond law

F = m * a

the net force acted on bullet , Fnet = 0.015 * 422500

Fnet = 6337.5 N

the net force acted on bullet is 6337.5 N

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