Question

A 1.0 × 10^3 kg sports car is initially traveling at 15 m/s. The driver then applies the brakes for several seconds so that -25 kJ of net work is done on the car. Calculate the initial and final kinetic energy of the car.

Answer 1

Answer:

Ea = 112500[J]

Eb = 87500[J]

Explanation:

To solve this problem we must use the principle of energy conservation which tells us that the energy of a body plus the work done or applied by the body equals the final energy of a body.

This can be easily visualized by the following equation:

$E_{A}+E_{friction}=E_{B}$

Now we must define the energies at points A & B.

For point A

At point A we only have kinetic energy since it moves at 15 [m/s]

So the kinetic energy

$E_{A}=\frac{1}{2}*m*v_{A}^{2} \\E_{A}=\frac{1}{2} *1000*(15)^{2} \\E_{A}=112500[J]$

The final kinetic energy can be calculated as follows:

$112500-25000=E_{B}\\E_{B}=87500[J]$