Question

A 1200.0-kg car speeds up from 16.0 to 20.0 m/s. How much work was done on the car to increase its speed?

Answer 1

Question 24637

$$m = 12000 \, kg, \, v_2 = 20 \, m/s, \, v_1 = 16 \, m/s$$

Work done on the car is the difference of kinetic energies, when car had final and initial velocity. $A = -(T_2 - T_1) = \frac{-m}{2} (v_2^2 - v_1^2) = -86400J$ (we obtain negative sign, because work was done on car). So, the work done on car is $86400J$ .