An 8.0g bullet, moving at 400 m/s, goes through a stationary block of wood in 4.0 x 10^-4s, emerging at a speed of 100 m/s. (a) what average force did the wood exert on the bullet? (b) how thick is the wood?

Explanation:

Initial velocity (u) of the Bullet = 400 m/sec

Final velocity (v) of the Bullet = 100 m/sec

Bullet passed through the block in (t) = 0.0004 sec  

Using 1st Equation of motion :

400 m/s = 100 m/s - a (0.0004)

Deceleration of Bullet = 750,000 m/sec^2

(a) F (force exerted by the wooden block on the bullet) = F (force exerted by the bullet on the wooden block)

F = m * a = 0.008 * 750,000 = 6000 N

(b) Using 3rd Equation of motion :

10000 = 160000 - 2 * 750,000 * S

Thickness of wood (S) = 0.1 m

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation:

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.