Question

A 30 g bullet travelling initially at 500 m/s penetrates 12 cm in to wooden block. The average force exerted wiII be:

• A. 31250 N
• B. 41250 N
• C. 31750 N
• D. 30450 N

Answer 1

The correct option is A 31250 N

K.E. of the bullet $=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times \frac{30}{1000}\times \left(500\right)=3750J$

Using the work energy theorem

Work done by retarding force =3750 J

$F_{retarding}\times distance=3750J$

$F_{retarding}=\frac{3750}{\frac{12}{100}}=31250N$