Question

bullet of mass 0.04 kg moving with a speed of 90 m s-1 enters a heavy wonden block and is stopped after a distance of 60 cm what is the average resistive force exerted by the block on the bullet

Answer 1

Answer:

Answer is 270N

Explanation:

bcoz we are given that:

           u=90

           v=0

           s=0.6m

So we have v² -u²=2as          

therefore a=  -u²/2s

so 90*90/2*0.6=6750

& f=m*a

∴ f= 0.04*6750

so f=270n

Answer 2

Answer: -270N

Explanation:

Given

m=0.04kg

u=90m/s

v= 0

s= 60cm=0.6m

from 3rd equation of motion

v^2=u^2+2as

0^2=90^2+2(0.6)a

0-1800=1.2a

-1800/1.2=a

a=-6750

F=ma

F=0.04×-6750

F= -270N

-ve sign indicate resistive or retarded force