Question

A 12 gram bullet is accelerated from rest to 7.0X102 m/s as it travels down a 20. cm yun barrel. Pasume constant acceleration. What was the accelerating force

Answer 1

Remember the following formula that relates the initial and final velocites of an object moving along a distance d at a constant acceleration a:

$v^2_f=v^2_0+2ad$

Since the bullet starts at rest, then the intial velocity is 0:

$\Rightarrow v^2_f=2ad$

Isolate a from the equation:

$\Rightarrow a=\frac{v^2_f_{}}{2d}$

Substitute v_f=70x10^2 m/s and d=20cm:

$\begin{gathered} a=\frac{(7.0\times10^2\frac{m}{s})^2}{2(20\operatorname{cm})} \\ =\frac{(7.0\times10^2\frac{m}{s})^2}{2(20\times10^{-2}m)} \\ =1.225\times10^6\cdot\frac{m}{s^2} \end{gathered}$

The net force is equal to the mass of the object times its acceleration, according to the Newton's Second Law of Motion:

$F=ma$

Substitute the value of the acceleration that was found previously, and m=12g:

$\begin{gathered} F=(12g)(1.225\times10^6\frac{m}{s^2}) \\ =(12\times10^{-3}kg)(1.225\times10^6\frac{m}{s^2}) \\ =14,700N \end{gathered}$

Therefore, the value of the acceleration force, was:

$14.7\times10^3N$