A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tree's trunk. What is the initial kinetic energy of the bullet? 153.125 J Correct: Your answer is correct. How much work does the tree do in stopping the bullet?

Answer: Work done = 153.125Joules, Work done = 0.003Nm

Explanation:

Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

Mathematically,

K. E = 1/2MV²

Where;

M = mass of the body = 2.5g = 0.0025kg

V = velocity of the body = 350m/s

Substituting this values in the formula, we have;

K. E = 1/2 * 0.0025*350²

K. E = 153.125Joules

Work done is the force applied to body to cause it to move through a distance.

Work = Force * distance

Force = ma = 0.0025 * 10

Force = 0.025N

Distance = 12cm = 0.12m

Work = 0.025*0.12

Work = 0.003Nm

work done by the tree in stopping the bullet is 0.003N

The work done by the tree in stopping the bullet is 0.00294 J

Explanation:

To determine how much work done by the tree in stopping the bullet, we first determine the force of the bullet.

Force = mass x acceleration due to gravity

Force = 0.0025 x 9.8 = 0.0245 N

Work done = Force x distance

= 0.0245 x 0.12

= 0.00294 J

Therefore, the work done by the tree in stopping the bullet is 0.00294 J