A) 140 J, 1846 N
B) 240 J, 184.6 N
C) 240 J, 1846 N
D) 240 J, 1746 N
Correct Answer: C
Solution :
The kinetic energy of the bullet on leaving the barrel is\[{{K}_{f}}=\frac{1}{2}m{{v}^{2}}\]\[=\frac{1}{2}(0.003){{(400)}^{2}}=240\text{J}\] The work done on the bullet is equal to the change in its kinetic energy. \[W=F\times ={{K}_{f}}-{{K}_{p}}\] where F is the average force exerted on the bullet. Thus, \[F=\frac{{{K}_{f}}-{{K}_{i}}}{x}\] \[=\frac{240-0}{0.13}=1846N\] Initial kinetic energy is zero, since the bullet was at rest initially.
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