A 45 g bullet, with a horizontal velocity of 498 m/s, stops 17 cm within a solid wall.?

Question

) What is the change in its mechanical energy?

What is the magnitude of the average force from the wall stopping it?

Answer 1

It had a kinetic energy (1/2 mv^2) and came to rest (KE=0).

Thus it lost 1/2 mv^2 in energy = 0.5*0.045*(498^2) = 5580 J

It goes from a speed u = 498 m/s to a speed v of 0 m/s in a distance s = .17 m.

v^2 = u^2 + 2as
-2as = u^2
a = -u^2/(2s)
a = -(498^2)/(2*0.17) = -729423 m/s^2

The force is F = ma = (729423)(0.045) = 32,800 N

Answer 2

a) 5,580.9 Joules.
b) 32824 Newtons

HFS a bullet that weighs 45 GRAMS? That's like a 0.50BMG round!