Question

a

Answer 1

@Michele_Laino

Answer 2

is the initial speed of the bullet:
3*104=312 m/sec?

Answer 3

oops..312 cm/sec?

Answer 4

Here, the work done by the friction force, has to be equal to the kinetic energy of the bullet, so we can write this:
\[\Large F \times d = \frac{1}{2}mv_0^2\]
where F is the requested force, d= 0.05 meters, and m=0.002 Kg, v_0=3.12 m/sec

Answer 5

so, dividing both sides by d, we get:
\[\Large F = \frac{{mv_0^2}}{{2d}} = \frac{{0.002 \times {{3.12}^2}}}{{2 \times 0.05}} = ...Newtons\]

Answer 6

that's right!