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M 150kg v 15ms KE ½ 150kg 15 ms2 KE ½ 150kg225 KE 16875J 2.
Example of kinetic energy equation. Using algebra you can solve for the unknown variable by rearranging all of the known variables to one side of the equation. 600 ½ m 20 2. KE ½ m v2.
E k 1 2 mv2 05 x 15x104 x 52002 2028 x 1011J PTO for rest of solution. KE is the kinetic energy in joules J. In other words any object body or particle with a mass and speed has kinetic energy.
Kinetic Energy Practice Problems 1. Velocity 20 ms. Compute its Kinetic energy.
A proton m 167 x 10-27 kg travels at a speed v 09900c 2968 x 10 8 ms. KE ½ mv2 KE. What is its kinetic energy.
Ek 12 10 kg 10kmh Ek 50 Joules. Also m 10 kg and v 10 kmh. Hence the kinetic energy in this case is 50 Joules.
500 ½ m 400. 500 ½ m 100. E k 100000 Joules or 1 x 10 5 Joules This example will guide you to calculate the kinetic energy manually.
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