Kinetic Energy

1. Kinetic Energy April 14th, 2011

2. What is Kinetic Energy • Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. • There are many forms of kinetic energy: • vibrational - the energy due to vibrational motion), • rotational (the energy due to rotational motion) • translational (the energy due to motion from one location to another)

3. Cont. • The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) that an object has depends upon two variables: •  the mass (m) of the object and the speed (v) of the object

4. Cont. • The following equation is used to represent the kinetic energy (KE) of an object: KE = (½)mv2 m = mass of object (kg) v = speed of object (m/s) • This equation reveals that the kinetic energy of an object is directly proportional to the square of its velocity. That means that for a double in velocity, the kinetic energy will increase by a factor of four

5. KE cont. • Kinetic energy is a scalar quantity; it does not have a direction. • Unlike velocity, acceleration, force, and momentum, the kinetic energy of an object is completely described by magnitude alone. • Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule.

6. Example 1. • Determine the kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s. EK = ½ mv2 = ½ (1000kg)(20.0m/s)2 = 200 000 Joules = 2.00 x 105 Joules or 200 kJ

7. Example 2. • If the roller coaster car in the above problem were moving with twice the speed, then what would be its new kinetic energy? EK = ½ mv2 = ½ (1000kg)(40.0m/s)2 = 800 000 Joules = 8.00 x 105 Joules or 800 kJ

8. Example 3. • Missy Diwater, the former platform diver for the Ringling Brother's Circus had a kinetic energy of 15kJ just prior to hitting the bucket of water. If Missy's mass is 50 kg, then what is her speed? EK = ½ mv2 15 000 J = ½ (50kg)v2 2(15000) = 50v2 30000 = v2 50 v =24.5m/s

9. Work-energy theorem • The net work done on an object is equal to its change in kinetic energy. • Note that the work in the work energy theorem (from yesterday’s class) is the work done on an object by a net force – it is the algebraic sum of work done by all forces. W = Ekf – Eki = ∆Ek • *** So the change in kinetic energy is equal to the work done on or by an object***

10. Example 4. • Calculate the velocity of a fist with a mass of 750g while being slammed into a board with a force of 50 N over a distance of 35 cm.(watch out for units!) EK = W = ½ mv2 W = F d = ½ (0.75kg) v2 50 N x 0.35 m = ½ (0.75kg) v2 17. 5/ 0.375 = v2 6.83 m/s = v

11. Example 5. • A shotputter heaves a 7.26kg shot with a final speed of 7.50m/s. a. What is the kinetic energy of the shot? b. The shot was initially at rest, how much work was done on it to give it this kinetic energy? a. Ek= ½ mv2 = ½ (7.26kg)(7.50m/s)2 = 204 J b. W = ∆Ek = Ekf – Eki = 204J – 0J = 204J

12. Homework • Pg. 238 #’s 19,20,21 • Pg. 245 #’s 22,23,25