A small truck has a mass of 2085 kg. How much work is required to decrease the speed of the vehicle from 22.0 m/s to 13.0 m/s on a level road?

Work required is 328387.5 Joules.

Explanation:

Given the following data;

Mass = 2085kg

Initial velocity, Vi = 13m/s

Final velocity, Vf =22m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

Where;

W is the work required.

Kf is the final kinetic energy possessed by the object.

Ki is the initial kinetic energy possessed by the object.

But Kinetic energy = ½MV²

W = ½MVf² - ½MVi²

Substituting into the equation, we have;

W = ½(2085)*22² - ½(2085)*13²

Simplifying the equation, we have;

W = 1042.5 * 484 - 1042.5 * 169

W = 504570 - 176182.5

W = 328387.5J

Therefore, the work required to decrease the speed of the vehicle is 328387.5 Joules.

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop