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.

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:

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

The final kinetic energy can be calculated as follows:

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.

i think its c if i'm wrong