A 3g bullet traveling at 350m/s hits a tree and slows down uniformly to a stop while penetrating a distance -?

Question

of 12cm into the tree's trunk. What was the force exerted on the bullet in bringing it to rest?

Answer 1

Since it slows down and stops, the tree does work on the bullet.

Work = W = Force * distance

Work = change in kinetic energy

So change in KE = 1/2mv^2 and Work = F x

F x = 1/2mv^2
F = (mv^2)/(2x) = 0.003 kg*350^2 m^2/s^2/(2*0.12 m)
F = 1531.3 kg m/s^2 = 1531.2 Nts

Answer 2

Kinetic capability (additionally entire) E of the Bullet till now hitting the tree is: E= (a million/2) m v^2, the place m= 3g =3*10^(-3) kg and v = 350m/s. So capability E =(a million/2) 3*10^(-3) (350)^2 Joules =183.seventy 5 Joules. Now if F Newtons is the (consistent) stress which the tree exerts on the bullet mutually because it particularly is penetrating in it.Then, the paintings W performed by ability of this stress is: W = F*s , the place s = 12 cm = 0.12m. So: W = 0.12 F Joules.This paintings is comparable to the kinetic capability of the bullet.(paintings required to deliver the bullet to relax or its kinetic capability to 0). So: 0.12 F =183.seventy 5 Or F=1531.25 N

Answer 3

s = 12 cm = 0.12 m
u = 350 m/s
v = 0 m/s

v^2 - u^2 = 2as
Put in the values :
a = - 510416.67 m/s^2

m = 0.003 kg

F = ma = 1531.25 N

Hope this helps.

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