A truck of mass 1000 kg increases its speed from 36 km h^-1 to 72 km h^-1. Find the increase in its kinetic energy.

Given:

Mass (m) = 1000 kg

1 km h^-1 = $\dfrac{\text{5}}{\text{18}}$ m s^-1

36 km h^-1 = $\dfrac{\text{5}}{\text{18}}$ x 36 = 10 m s^-1

So, initial speed (v₁) = 10 m s^-1

1 km h^-1 = $\dfrac{\text{5}}{\text{18}}$ m s^-1

72 km h^-1 = $\dfrac{\text{5}}{\text{18}}$ x 72 = 20 m s^-1

So, final speed (v₂) = 72 km h^-1 = 20 m s^-1

Increase in its kinetic energy = ?

Increase in kinetic energy = $\dfrac{\text{1}}{\text{2}}$ m[(v₂)² - (v₁)²]

= $\dfrac{\text{1}}{\text{2}}$ x 1000 x [(20)² - (10)²]

= 500 x [400 - 100]

= 500 x 300

= 150000 J or 1.5 x 10⁵ J

So increase in kinetic energy = 1.5 x 10⁵ J.