Physics Energy and Society

Question

A 2.0 g bullet initially moving with a speed of 87 m/s passes through a block of wood. On exiting the block of wood, the bullet’s velocity is 12 m/s, (the change in height is negligible). 

a. How much WORK did the force of friction do on the bullet as it passes through the block? 

b. If the block was 4.0 cm thick, what was the average force that the wood exerted on the bullet? 

Answer 1

The work done by the friction of the wooden block is in this case the conversion of the kinetic energy of the bullet to heat, deformation,... of the wood and bullet.

So we can calculate this work done by observing the change in kinetic energy of the bullet.

a) W = ΔKE = 0.5*m*Δ(v²) = 0.5* 2.0*10^-3 kg * (12² - 87²) m²/s² = -7.425 J (~ -7.4 J, 2 SF)

The work is negative, since the friction slows the bullet down.

Work is done when a force acts on an object and that object undergoes a displacement.

Since we are only interested in the average force (~ treating it as a constant force), the calculations are simplified: W = F Δx cos(α), with α the angle between the force vector and the displacement vector (in this case α = 180°, friction opposes motion and thus displacement).

b) F = W / (Δx*cos(α)) = 7.425 J / (4.0*10^-2 m) = 185.625 N (~ 1.9*10 N, 2 SF)

The friction does -7.4 J of work on the bullet, meaning that if the wooden block was 4.0 cm thick, the magnitude of its average force on the bullet was 1.9*10 N.