Question

what average force is necessary to stop a bullet of mass 20gm and speed of 250m/sec as it penetrates wood to a distances of 12 cm

Answer 1

using kinematics eq.
v^2 =u^2 - 2as
0 = (250 )^2 - 2*0.12*a
0.24a=62500
a= 62500/0.24
now
F = ma
= 20/1000 * 62500/0.24
solve it
F = 5208 N

Answer 2

Answer:

The average force necessary to stop a bullet is 5208 N.

The average force necessary to stop bullet with given specification while penetrating the wood, will be calculated as follows –

Given is the distance of wood d = 12cm = 0.12 m

Initial velocity u = 250 m/ s

Final velocity v = 0 (as bullet stops)

Mass of the bullet m = 20 gm = 20 / 1000 kg

From the equation of motion, we have -  

$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{a} \mathrm{s}$

$\mathrm{V}^{2}=\mathrm{u}^{2}+2 \mathrm{a} \mathrm{d}$

0 x 0 = 250 x 250 + 2 x a x 0.12

a = 62500 / 0.24

$a=62500 / 0.24 \mathrm{m} / \mathrm{s}^{2}$

Now as we know that Force = mass x acceleration  

Force required to stop the bullet = mass of bullet x acceleration of bullet

F = m  x a

F = 20 / 1000 x 62500 / 0.24

F = 5208 N