Work, Energy, and Power - Work-Energy Theorem

A block of mass $m = 2\text{ kg}$ is moving on a frictionless horizontal surface with an initial velocity of $u = 5\text{ m/s}$. A constant force is applied on the block in the direction of motion, increasing its velocity to $v = 10\text{ m/s}$. Calculate the work done by the force.

$W = \Delta K = \frac{1}{2}m(v^2 - u^2)$ $W = \frac{1}{2} \times 2 \times (10^2 - 5^2)$ $W = 1 \times (100 - 25)$ $W = 75\text{ J}$

According to the Work-Energy Theorem, the work done by the external force is equal to the change in kinetic energy. Since the mass and both velocities are known, we calculate the difference between final and initial kinetic energies.

A bullet of mass $20\text{ g}$ ($0.02\text{ kg}$) moving with a speed of $100\text{ m/s}$ enters a heavy wooden block and is stopped after a distance of $50\text{ cm}$ ($0.5\text{ m}$). What is the average resistive force exerted by the block on the bullet?

$W_{net} = \Delta K$ Since the bullet stops, $v_f = 0$. $F \cdot d \cdot \cos(180^\circ) = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$ $-F \cdot 0.5 = 0 - \frac{1}{2}(0.02)(100)^2$ $-0.5F = -0.01 \times 10000$ $-0.5F = -100$ $F = \frac{100}{0.5} = 200\text{ N}$

The work done by the resistive force is negative because the force is opposite to the displacement. By setting the work done ($F \cdot d$) equal to the change in kinetic energy, we can solve for the unknown resistive force $F$.