Question.
Calculate the work required to be done to stop a car of $1500 \mathrm{~kg}$ moving at a velocity of $60 \mathrm{kmh}^{-1}$.
Calculate the work required to be done to stop a car of $1500 \mathrm{~kg}$ moving at a velocity of $60 \mathrm{kmh}^{-1}$.
Solution:
Given, mass of car, $\mathrm{m}=1500 \mathrm{~kg}$;
velocity of car, $\mathrm{v}=60 \mathrm{kmh}^{-1}=60 \times \frac{5}{18}=\frac{50}{3} \mathrm{~m} / \mathrm{s}$
Work done $=$ Change in kinetic energy of the car
or $W=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=\frac{1}{2} m\left(v^{2}-u^{2}\right)$
or $W=\frac{1}{2}(1500)\left[(0)^{2}-(50 / 3)^{2}\right]$
or $W=-\frac{1}{2} \times 1500 \times \frac{2500}{9}=-208333 J$
Given, mass of car, $\mathrm{m}=1500 \mathrm{~kg}$;
velocity of car, $\mathrm{v}=60 \mathrm{kmh}^{-1}=60 \times \frac{5}{18}=\frac{50}{3} \mathrm{~m} / \mathrm{s}$
Work done $=$ Change in kinetic energy of the car
or $W=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=\frac{1}{2} m\left(v^{2}-u^{2}\right)$
or $W=\frac{1}{2}(1500)\left[(0)^{2}-(50 / 3)^{2}\right]$
or $W=-\frac{1}{2} \times 1500 \times \frac{2500}{9}=-208333 J$