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?
Question
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]