Question

A bullet of mass 10 g leaves a rifle at an initial velocity of 1000 m/s and strikes the earth at the same level with a velocity of 500 m/s. The work done in joule overcoming the resistance of air will be

• A. 375
• B. 3750
• C. 5000
• D. 500

Answer 1

The Correct Option is B

To solve this problem, we need to apply the concept of kinetic energy and the work-energy principle. The work done is equal to the change in kinetic energy of the bullet.

1. First, calculate the initial kinetic energy (KE_initial) of the bullet when it leaves the rifle.
   • Mass of the bullet, m = 10 \, \text{g} = 0.01 \, \text{kg} (since 1 g = 0.001 kg).
   • Initial velocity, v_1 = 1000 \, \text{m/s}.

   The kinetic energy formula is:

   KE = \frac{1}{2}mv^2

   Thus,

   KE_{\text{initial}} = \frac{1}{2} \times 0.01 \times (1000)^2 = 5 \times 10^3 \, \text{Joules}

2. Next, calculate the final kinetic energy (KE_final) of the bullet when it strikes the earth.
   • Final velocity, v_2 = 500 \, \text{m/s}.

   Thus,

   KE_{\text{final}} = \frac{1}{2} \times 0.01 \times (500)^2 = 1.25 \times 10^3 \, \text{Joules}

3. The work done by the resistance of air is the change in kinetic energy, which is the difference between the initial and final kinetic energies:

   W = KE_{\text{initial}} - KE_{\text{final}}

   Therefore,

   W = 5000 - 1250 = 3750 \, \text{Joules}

Thus, the work done in overcoming the resistance of air is 3750 Joules, which corresponds to the correct answer.