Work, Energy and Simple Machines - The Work-Energy Theorem

A car of mass $1500 \text{ kg}$ is moving with a velocity of $10 \text{ m/s}$. The driver applies the accelerator to increase the velocity to $20 \text{ m/s}$. Calculate the work done by the engine.

Given: mass $m = 1500 \text{ kg}$, initial velocity $u = 10 \text{ m/s}$, final velocity $v = 20 \text{ m/s}$. Step 1: Calculate initial kinetic energy: $E_{ki} = \frac{1}{2} \times 1500 \times 10^2 = 750 \times 100 = 75000 \text{ J}$ Step 2: Calculate final kinetic energy: $E_{kf} = \frac{1}{2} \times 1500 \times 20^2 = 750 \times 400 = 300000 \text{ J}$ Step 3: Calculate Work Done ($W = E_{kf} - E_{ki}$): $\begin{array}{r} 300000 \\ -75000 \\ \hline 225000 \end{array}$ Work done = $225000 \text{ J}$ or $225 \text{ kJ}$.

According to the Work-Energy Theorem, the work done by the engine results in an increase in the car's kinetic energy. By subtracting the initial energy from the final energy, we find the total work performed.

How much work is required to stop a moving ball of mass $0.5 \text{ kg}$ traveling at a speed of $4 \text{ m/s}$?

Given: $m = 0.5 \text{ kg}$, $u = 4 \text{ m/s}$, $v = 0 \text{ m/s}$ (since it stops). Using the theorem: $W = \frac{1}{2}m(v^2 - u^2)$ $W = \frac{1}{2} \times 0.5 \times (0^2 - 4^2)$ $W = 0.25 \times (0 - 16)$ $W = -4 \text{ J}$

The work done is $-4 \text{ J}$. The negative sign indicates that the work is done in the direction opposite to the motion (opposing force) to reduce the kinetic energy to zero.