Answer to Question #145722 in Physics for Justin Allen

Question

If a car of mass 400 kg reduces speed from 80 km/h to 34 km/h in 17s. Calculate the braking power required to give this change of speed.

Answer 1

Let's first convert km/h to m/s:

$$v_i=(80\ \dfrac{km}{h})\cdot (\dfrac{1000\ m}{1\ km})\cdot(\dfrac{1\ h}{3600\ s})=22.22\ \dfrac{m}{s},$$ $$v_f=(34\ \dfrac{km}{h})\cdot (\dfrac{1000\ m}{1\ km})\cdot(\dfrac{1\ h}{3600\ s})=9.44\ \dfrac{m}{s}.$$

We can find the braking power required to give this change of speed from the definition of the power:

$$P=\dfrac{W}{t},$$

here, $P$ is the braking power, $W$ is the work done by the braking force to change the speed from 80 km/h to 34 km/h, $t$ is time.

We can find the work done by the braking force from the Work-Kinetic Energy Theorem:

$$KE_f-KE_i=-W,$$ $$W=-(\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2)=-\dfrac{1}{2}m(v_f^2-v_i^2),$$ $$W=-\dfrac{1}{2}\cdot\ 400\ kg\ \cdot((9.44\ \dfrac{m}{s})^2-(22.22\ \dfrac{m}{s})^2)=80923\ J.$$

Finally, substituting $W$ into the formula for the power we can calculate the braking power required to give this change of speed:

$$P=\dfrac{W}{t}=\dfrac{80923\ J}{17\ s}=4760\ W=4.76\ kW.$$

Answer:

$P=4760\ W=4.76\ kW.$