Question

A bullet of mass 4 into ten power minus 2 moving with a velocity of 20m/s peries through sand bag and comes to rest after 0.5 sec the resistance offered by the sand bag is

Answer 1

Answer:

1.6 N

Explanation:

m = 4 x 10^-2 kg

u = 20 m/s

t = 0.5 sec

v = 0 m/s (rest)

thus using laws of kinematics

v = u + a*t

a = - 20/0.5

a   = - 40 m/sec²

Now,

Frictional force = m*a = 4 * 10⁻² * 40

                                 = 1.6 N