A bullet of mass \(0.10 \mathrm{~kg}\) moving with a speed of $100 \mathrm{~m} / \mathrm{s}\( enters a wooden block and is stopped after a distance of \)0.20 \mathrm{~m}$. What is the average resistive force exerted by the block on the bullet ? (A) \(2.5 \times 10^{2} \mathrm{~N}\) (B) \(25 \mathrm{~N}\) (C) \(25 \times 10^{2} \mathrm{~N}\) (D) \(2.5 \times 10^{4} \mathrm{~N}\)

In this problem, we know: - The mass of the bullet (m) is \(0.10 \mathrm{~kg}\). - The initial velocity of the bullet (v1) is \(100 \mathrm{~m}/\mathrm{s}\). - The final velocity of the bullet (v2) is \(0 \mathrm{~m}/\mathrm{s}\) (because the bullet stops). - The distance traveled by the bullet within the block (d) is \(0.20 \mathrm{~m}\). Our goal is to find the average resistive force exerted by the wooden block on the bullet, which we will denote as F.

The work-energy principle states that the work done by external forces on an object is equal to the change in its kinetic energy. So, we have: \(W = \Delta KE\) Since the bullet is stopped by the resistive force exerted by the block, we know that the work done by this force is equal to the change in kinetic energy of the bullet.

The kinetic energy (KE) of the bullet for both initial and final states can be calculated using the formula: \(KE = \frac{1}{2}mv^2\) Now, we calculate the change in kinetic energy: \(\Delta KE = KE_{final} - KE_{initial}\) Since the final velocity of the bullet is 0 m/s, its final kinetic energy is zero. Therefore: \(\Delta KE = 0 - KE_{initial}\) To find the initial kinetic energy, we can plug in the given values for mass and initial velocity: \(KE_{initial} =\frac{1}{2}(0.10 \mathrm{~kg})(100 \mathrm{~m}/\mathrm{s})^2 = 500 \mathrm{~J}\) Now, we can calculate the change in kinetic energy: \(\Delta KE = 0 - 500 \mathrm{~J} = -500 \mathrm{~J}\)

Now that we have the change in kinetic energy, we can use the work-energy principle to find the work done by the resistive force: \(W = \Delta KE = -500 \mathrm{~J}\)

Finally, we can find the average resistive force exerted by the wooden block using the formula: \(W = F \cdot d\) Let's solve for F: \(F = \frac{W}{d}\) Now we can plug in the values for work done and distance: \(F = \frac{-500 \mathrm{~J}}{0.20 \mathrm{~m}} = -2500 \mathrm{~N}\) Since the force is exerted in the opposite direction of the initial motion of the bullet, we have a negative sign indicating that it acts in the opposite direction. However, we are asked for the magnitude of the average resistive force, so we can take the absolute value: \(|F| = 2500 \mathrm{~N}\) Thus, the average resistive force exerted by the wooden block on the bullet is \(2.5 \times 10^2 \mathrm{~N}\). The correct answer is (A) \(2.5 \times 10^{2} \mathrm{~N}\).