Chapter 5 - Work and Energy - Learning Path Questions and Exercises - Exercises - Page 176: 30

a). $375kJ$ b). $375kJ$

a). m=1200kg, v=$90km/hr=90\times\frac{1000m}{3600s}=25m/s$ KE=$\frac{1}{2}\times1200\times25^{2}=375kJ$ b). To stop the car, the work to be done should be equal to the initial KE because the velocity becomes zero finally as is the KE. So, work to be done is $375kJ$.