Question

A bullet of mass .03 kg moving with a speed of 400m/s penetrates 12cm into a fixed block of wood calculate the average force exerted by the wood on the bullet.

Answer 1

V^2 = Vo^2 + 2a*d = 0.
400^2 + 2a*0.12 = 0,
160,000 + 0.24a = 0, a = -666,667 m/s^2.

F = M*a = 0.03 * (-666,667) = -20,000 N.

Answer 2

Can you please explain it in simple way .

Answer 3

K.E. of bullet = Workdone
K.E. = 1/2 mv^2
Workdone =Force* Distance
1/2mv^2 = Force *0.12
1/2*0.03*(400)^2 = Force *0.12
Force = 20kn