Question

A bullet of mass 0.01 kg strikes of wooden target 0.1 M thick with a velocity of 400 metre per second it emerges out from the target with a velocity of 100 metre per second calculate loss of kinetic energy and average force of friction of wood

Answer 1

M=0.01kg

U=400m/s

V=100m/s

1.Kinetic energy when the bullet was launched

Ke=1/2mv²

Ke=1/2*0.01*(400)²

Ke=1/2*1600

Ke=800J

Kinetic energy after the bullet emerges out

Ke=1/2mv²

Ke=1/2*0. 01*(100)²

Ke=1/2*100

Ke=50J

Loss of kinetic energy is,

Ke1-Ke2

=800-50

=750J

So, the loss of kinetic energy is 750J.

2. Thickness of wood=0.1m

=1cm

Mass=Desity/Volume

Mass=1/1

Mass=1kg

F=mg

F=1*10

F=10J

MARK ME AS THE BRAINLIEST PLEASE.