Question

A bullet of mass 25 gm and velocity 50 m/s passes through a board 7.5 cm thick and emerges with a velocity of 40m/s.what is the average force exerted by the boardon the bullet?

Answer 1

Work = Fd where F is force, d is distance

But work is also W = ΔEk where Ek is kinetic energy

In general, kinetic energy Ek = ½mv^2

So here, ΔEk = ½m(v1^2 – v2^2)
= ½(0.025)(50^2 – 40^2) = 11.25J

Then Fd = F(0.075) = 11.25, so F = 11.25/0.075 = 150N

The average force is 150N.

Answer 2

Answer:

150N

Explanation:

Applyng work-energy theorem

W=Chang in kinetic energy

_ f x s = 1/2 m ( v² -u²) by substituting

_ f x 0.075 = 1/2 (0.025) ( 40² -50²) solve for f

f = 150 N .