A bullet of mass 0.04 kg moving with a speed of 90m/s enters a heavy wooden block and is stopped after a distance of 60 cm . What is the average force exerted by the block on the bullet?
Given,m=0.04kg
u=90m/s
v=0m/s
S=60cm
From 3rd equation of motion,
2aS=v2−u2
a=−u22S
a=−90×902×60×10−2=−6750m/s2
The resistive force, f=ma
f=−0.04×67.5
f=−270N