A 5.0-g bullet leaves the muzzle of a rifle with a speed of 20 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.82-m-long barrel of the rifle?

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Answer 1 · 2017-04-25T16:57:01

The force is #=1.22N#

To find the acceleration in the barrel, we apply the equation

#v^2=u^2+2as#

#u=0#

The speed of the bullet leaving the muzzle is #v=20ms^-1#

Barrel length is #s=0.82m#

So,

The acceleration is

#a=v^2/(2s)=20^2/(2*0.82)=243.9ms^-2#

To find the force, we apply Newton's Second Law

#F=ma=0.005*243.9=1.22N#