Question A rifle can shoot a 4.2 g bullet at a speed of 965 m/s. If the work on the bullet is done over a distance of 0.75 m, what is the average force on the bullet?

Answer: The average force on the bullet will be “2607.43 N“. Explanation: The given values are: Mass of the bullet, m = 4.2 g Final speed, v = 965 m/s Initial speed, u = 0 Distance, S = 0.75 m According to the Energy theorem, ⇒ [tex]Work \ done = Change \ in \ kinetic \ energy[/tex] ⇒ [tex]F S = (\frac{1}{2} )m(v^2 – u^2)[/tex] On substituting the values, we get ⇒ [tex]F\times 0.75=0.5\times 0.0042 [(965)^2 – 0][/tex] ⇒ [tex]0.75 F = 0.5\times 0.0042\times 931225[/tex] ⇒ [tex]0.75F=1955.57[/tex] ⇒ [tex]F=\frac{1955.57}{0.75}[/tex] ⇒ [tex]=2607.43 \ N[/tex] Log in to Reply

Answer:The average force on the bullet will be “

2607.43 N“.Explanation:The given values are:

Mass of the bullet,m = 4.2 g

Final speed,v = 965 m/s

Initial speed,u = 0

Distance,S = 0.75 m

According to the Energy theorem,⇒ [tex]Work \ done = Change \ in \ kinetic \ energy[/tex]

⇒ [tex]F S = (\frac{1}{2} )m(v^2 – u^2)[/tex]

On substituting the values, we get⇒ [tex]F\times 0.75=0.5\times 0.0042 [(965)^2 – 0][/tex]

⇒ [tex]0.75 F = 0.5\times 0.0042\times 931225[/tex]

⇒ [tex]0.75F=1955.57[/tex]

⇒ [tex]F=\frac{1955.57}{0.75}[/tex]

⇒ [tex]=2607.43 \ N[/tex]