Work question (physics)

Question

Calculate the work done by a gun to accelerate a 5.0 g bullet to a speed of 1100 km/h from a gun with a 10.0 cm barrel.

Use this W=F d

Answer: (233J or 2.3 x 102 J)

Please show the process of finding this answer

Answer 1

Given:

m = 5.0 g = 5.0 * 10^-3 kg

v_f = 1100 km / hr ≈ 305.555...6

d = 10.0 cm = 0.1 m

Find:

W = ? J

Solution:

W = F * d

F = m * a

v_f² = v_i² + 2 * a * d

a = v_f² / 2 * d = ( 305.55...6 m / s )² / ( 2 * 0.1 m )

∴ a ≈ 466821 m / s²

F = m * a = ( 5.0 * 10^-3 kg ) * ( 466821 m / s² )

∴ F ≈ 2334 N

W = F * d = ( 2334 N ) * ( 0.1 m )

∴ W ≈ 233 J

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

Also, I don't know why it would give that much information unless your teacher wanted to see how much force that a gun's bullet undergoes and the MASSIVE acceleration it experiences to get up to that speed in such a short barrel.

You would arrive at the same, correct answer if you had equated that the potential work that would need to be done equals the final kinetic energy of the bullet.

W = 1/2 * m * v_f² = 0.5 * ( 5(10^-3) kg ) * ( 305.55...6 m / s )²

W = 233 J

But any who, that's how you get that answer thru the method asked in the problem.

Answer 2

Fd = mv²/2; F = mv²/(2d) = 0.005kg(1100·1000m/3600s)²/(2·0.1m) = 2334.1 N

W = mv²/2 = 0.005kg(1100·1000m/3600s)²/2 = 233.41 J