A 3.0-g bullet traveling at 350 m/s hits a tree and slows down uniformly to a stop while penetrating...?

Question

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

Answer 1

It's going from 350 m/s to 0 m/s in 12 cm (0.12 m)
The average speed, since it is decelerating uniformly, is 175 m/s. At 175 m/s, it will take 0.12 m / 175 m/s = 0.12/175 sec. to stop.
the deceleration is 350 m/s / (.12/175) sec. = 510416.67 m / sec./ sec.

The force, which you are asking about, is the acceleration (deceleration) multiplied by the mass of the object, 3 g = 0.003 kg.
The force is 0.003 kg * 350 m/s / (.12 /175)) sec.
Your answer is in kg * m / (sec.^2), or Newtons.
I get 1530.25 N, or 1500 N if you are concerned about significant digits. (The answer, realistically, can't be more precise than the question.)

Answer 2

Kinetic power (additionally entire) E of the Bullet in the previous hitting the tree is: E= (a million/2) m v^2, the place m= 3g =3*10^(-3) kg and v = 350m/s. So power E =(a million/2) 3*10^(-3) (350)^2 Joules =183.75 Joules. Now if F Newtons is the (consistent) rigidity which the tree exerts on the bullet at the same time because it quite is penetrating in it.Then, the paintings W completed with the aid of this rigidity is: W = F*s , the place s = 12 cm = 0.12m. So: W = 0.12 F Joules.This paintings is comparable to the kinetic power of the bullet.(paintings required to hold the bullet to sit down down back or its kinetic power to 0). So: 0.12 F =183.75 Or F=1531.25 N

Answer 3

Kinetic Energy Ke = Work done by force F slowing the bullet over distance s.
Ke=W=Fs
Ke=(1/2) mv^2
F=Ke/s

F=(1/2s) mv^2
F=(1/(2 x 0.12)) .003 (350)^2=1530 N