Answer to Question #285670 in Physics for dddd

Question

How much work is required to accelerate a 1500kg car from 15 m/s to 40 m/s?What is the average net force acting on the car if it covers a distance of 275m? 

Answer 1

Given:

$v_1=\rm 15\: m/s$

$v_2=\rm 40\: m/s$

$m=\rm 1500\: kg$

$d=\rm 275\: m$

The energy-work theorem says

$$\frac{mv_2^2}{2}-\frac{mv_1^2}{2}=W=Fd$$

(a) the work done

$$W=\frac{mv_2^2}{2}-\frac{mv_1^2}{2}\\=\frac{1500}{2}(40^2-15^2)=1\:031\:250\: \rm J$$

(b) the average force

$$F=\frac{W}{d}=\frac{1031250}{275}=3\:750\: \rm N$$