A$\mathbf{1700}\mathbf{}\mathbf{\text{kg}}$Buick moving atbrakes to a stop, at uniform deceleration and without skidding, over a distance of$\mathbf{93}\mathbf{}\mathbf{\text{m}}$. At what average rate is mechanical energy transferred to thermal energy in the brake system?

1. Mass of the Buick$m=1700\text{kg}$
2. Initialspeed of the car,$v_i=83\text{km/hr}$
3. Distance travelled by the car as it comes to stop,$x=93\text{m}$.

The driver of afast-moving vehicle applies brakes to make the vehiclecome to a stop. When it comes to a stop, its mechanical energy gets converted to thermal energy in the brake system. This heats the system. Thus, the conversion of energy is seen when the vehicle comes to a stop.

Formula:

Rate of conversion of energy that is power$P=\frac{dE}{dt}$,…(i)

Kinetic energy of the system$K.E.=\frac{1}{2}m{v}^2$,…(ii)

Third kinematic equation of motion$v_f^2=v_i^2+2a(x-x_0)$,…(iii)

Second kinematic equation of motion,$v_f=v_i+at$. …(iv).

The initial speed of the Buck is

$\begin{array}{c}{v}_i=\text{83}\frac{\text{km}}{\text{hr}}\times \frac{\text{1000m}}{\text{1km}}\text{×}\frac{\text{1hr}}{\text{3600s}}\\ =\text{23.056}\frac{\text{m}}{\text{s}}\end{array}$

Now, the mechanical energy (M.E.) of the Buick is equal to the kinetic energy. This is calculated using equation (ii) as

$\begin{array}{c}ME=\frac{1}{2}\times 1700\text{kg}\times {\left(23.056\frac{\text{m}}{\text{s}}\right)}^2\\ =4.5184\times {10}^5\text{J}\\ \approx 4.52\times {10}^5\text{J}\end{array}$

Next, we determine the deceleration of the car as per equation (iii). This is given by

$\begin{array}{c}0={\left(23.056\frac{\text{m}}{\text{s}}\right)}^2+2\left(a\right)(93\text{m}-0)\\ a=\frac{{\left(23.056\frac{\text{m}}{\text{s}}\right)}^2}{-2\times 93\text{m}}\\ =-2.85795\frac{\text{m}}{{\text{s}}^{\text{2}}}\\ \approx -2.86{\text{m/s}}^{\text{2}}\end{array}$

The negative sign indicates that it is a deceleration.

Now we need to determine the time taken by the car to come to rest to determine the rate of energy transfer.

Using equation (iv) and the given values, we get the time taken as

$\begin{array}{c}0=23.056\text{\hspace{0.17em}m/s}-2.85795{\text{m/s}}^{\text{2}}t\\ t=\frac{23.05}{2.86}\text{s}\\ =8.067\text{s}\end{array}$

Now, the rate of energy transfer using equation (i) is given by

$\begin{array}{c}P=\frac{4.52\times {10}^5\text{\hspace{0.17em}J}}{8.067\text{\hspace{0.17em}s}}\\ =5.6\times {10}^4\text{W}\end{array}$

Hence, the value of the average rate of energy is$5.6\times {10}^4\text{W}$.