A 25 g bullet with a muzzle velocity of 300 m/s is fired into a board 4.0 cm thick. For a soft board, the bullet goes through the board and emerges with a speed of 50 m/s. What was the average force exerted on the bullet by the board?

Answer= 27344 N
Show Work

V^2 = Vo^2 + 2a*d,

a = (V^2-Vo^2)/2d = (50^2-300^2)/0.08 = -1,093,750 m/s^2.

F = M*a = 0.025 * (-1,093,750) =

To find the average force exerted on the bullet by the board, we can use the principle of work and energy.

First, let's calculate the initial kinetic energy (KE_initial) of the bullet using the mass (m) and the muzzle velocity (v):

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

Plugging in the values:
KE_initial = (1/2) * 0.025 kg * (300 m/s)^2 = 1125 J

Next, let's calculate the final kinetic energy (KE_final) of the bullet using the mass (m) and the final velocity (v):

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

Plugging in the values:
KE_final = (1/2) * 0.025 kg * (50 m/s)^2 = 31.25 J

Since work (W) is equal to the change in kinetic energy, we can find the work done on the bullet by the board by subtracting the final kinetic energy from the initial kinetic energy:

W = KE_initial - KE_final

W = 1125 J - 31.25 J = 1093.75 J

Finally, we can calculate the average force (F_avg) exerted on the bullet using the work done and the distance (d) traveled through the board:

F_avg = W / d

Here, the distance the bullet traveled through the board is 0.04 m (4.0 cm = 0.04 m).

Plugging in the values:
F_avg = 1093.75 J / 0.04 m = 27343.75 N (rounded to 27344 N)

Therefore, the average force exerted on the bullet by the board is 27344 N.