Question

a 1900 kg car slows down from a speed of 25m/s to a speed of 15m/s in 6.25s. how much work was done on the car

Answer 1

`-3.8\cdot 10^5 J`

Step-by-step explanation:

According to the work-energy theorem, the work done on the car is equal to its variation of kinetic energy:

`W=\Delta K=K_f -K_i = (1)/(2)mv_f^2-(1)/(2)mv_i^2`

where

m = 1900 kg is the mass of the car

vf = 15 m/s is the final speed

vi = 25 m/s is the initial speed

Substituting the data, we find

`W=(1)/(2)(1900 kg)(15 m/s)^2-(1)/(2)(1900 kg)(25 m/s)^2 =-3.8\cdot 10^5 J`

and the work is negative because it is done against the motion of the car (in fact, the car slows down, so it loses kinetic energy).