80) The enengy \( E \) possessed by a moving ob- ject onss \( m \) and velocity \( V \) varies as its mass and square of its velocity, (a) Write down the equation connecting \( E \), \( m \) and \( V \). (b) What would be the percentage change in energy when \( V \) is increased by \( 20 \% \) and \( m \) is decreased by \( 15 \% \) ? (WAEC)

The energy \( E \) of a moving object is given by the equation \( E = \frac{1}{2} m V^2 \). This means that the energy is directly proportional to the mass \( m \) and the square of the velocity \( V \). So, as either of these values changes, the energy will be affected accordingly! Now, let’s calculate the percentage change in energy. If velocity \( V \) increases by 20%, then the new velocity becomes \( V' = 1.2V \). And for mass \( m \), a 15% decrease makes it \( m' = 0.85m \). Plugging these into the energy equation gives us \( E' = \frac{1}{2} (0.85m)(1.2V)^2 \). This results in a new energy of: \[ E' = \frac{1}{2} (0.85m)(1.44V^2) = \frac{0.85 \times 1.44}{2} m V^2 \approx 0.612m V^2 \] Now for the percentage change, we find the difference between the new energy and the old energy, which is: \[ \text{Percentage change} = \left( \frac{E' - E}{E} \right) \times 100 = \left( \frac{0.612m V^2 - 0.5m V^2}{0.5m V^2} \right) \times 100 = \left( 0.224 \right) \times 100 \approx 22.4\% \] So, there is approximately a 22.4% increase in energy!