A 5.00 gram bullet has a velocity of 300 m/s. It strikes a block of wood and penetrates to a depth of 2.00 cm. The average force of the block of wood on the bullet is
A) 9,340 N.
B) 11,250 N.
C) 15,200 N.
D) 19,500 N.
E) 21,400 N.

Question

Quizplus · Accepted Answer

To find the average force exerted by the wood on the bullet, we can use the work-energy principle. The work done by the force is equal to the change in kinetic energy of the bullet. The kinetic energy (KE) of the bullet before it stops is given by $ KE = \frac{1}{2}mv^2 $, where $ m = 0.005 $ kg (5.00 grams converted to kilograms) and $ v = 300 $ m/s. The work done by the force is also equal to the force times the distance over which it acts ( $ W = Fd $ ), where $ d = 0.02 $ m (2.00 cm converted to meters).First, calculate the initial kinetic energy of the bullet:$$ KE = \frac{1}{2} \times 0.005 \times 300^2 = \frac{1}{2} \times 0.005 \times 90000 = 225 $$ Joules.Since all this energy is used to do work against the force of the wood (and assuming no energy is lost to other forms like heat, sound, etc., for simplicity), we set the work done equal to this kinetic energy:$$ W = Fd = 225 $$ Joules.Solving for $ F $ gives:$$ F = \frac{W}{d} = \frac{225}{0.02} = 11250 $$ Newtons.Therefore, the average force of the block of wood on the bullet is 11,250 N.

A 500 Gram Bullet Has a Velocity of 300 M/s