APRIL 14 TH, 2011 Kinetic Energy

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)

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

Cont. The following equation is used to represent the kinetic energy (KE) of an object: KE = (½)mv 2 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

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.

Example 1. Determine the kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s. E K = ½ mv 2 = ½ (1000kg)(20.0m/s) 2 = Joules = 2.00 x 10 5 Joules or 200 kJ

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? E K = ½ mv 2 = ½ (1000kg)(40.0m/s) 2 = Joules = 8.00 x 10 5 Joules or 800 kJ

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? E K = ½ mv J = ½ (50kg)v 2 2(15000) = 50v = v 2 50 v =24.5m/s

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 = E kf – E ki = ∆E k *** So the change in kinetic energy is equal to the work done on or by an object***

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!) E K = W = ½ mv 2 W = F d = ½ (0.75kg) v 2 50 N x 0.35 m = ½ (0.75kg) v / = v m/s = v

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. E k = ½ mv 2 = ½ (7.26kg)(7.50m/s) 2 = 204 J b. W = ∆E k = E kf – E ki = 204J – 0J = 204J

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