Energy 3U/3UB Physics

Energy Energy is a scalar quantity measured in Joules (J). 1 J = 1 kg m2/s2 or 1 Nm Energy is a measure of the ability of an object to do work. If work is done ON an object it gains energy and is a positive value. If work is done BY an object it loses energy and is a negative value.

Energy Types Work – the energy pathway taken to change an amount of one type of energy and/or to change the type of energy. W = Fd The Work-Energy Theorem states that the work done is equal to the change in energy (type or amount). W = ∆E Kinetic energy is the energy of motion, an object of mass with velocity. Ek = ½ mv2

Energy Types Potential energy refers to the ability to do work – usually a stored energy. There are many types: Gravitational Potential Energy – energy due to position above a gravitational body: Eg = mgh Elastic Potential Energy – energy stored in stretched materials that have the ability to restore back to their original shape Chemical Potential Energy – stored in molecular bonds and released when broken or combined with other elements (combustion, food) Nuclear Potential Energy – energy released when atoms undergo fission or fusion

Energy Types Mechanical Energy – the sum of the kinetic and gravitational potential energy. Law of Conservation of Mechanical energy states the energy before an event is equal to the energy afterwards (ignoring dissipative forces).

Energy changes/conservation of energy Example 1) An object is lifted up You do work to lift the object up. If you used 20.0 J to do this, you lost 20.0 J while the object gained 20.0 J of Gravitational Potential energy. Work done by you = - 20.0 J, Work done on object = 20.0 J Example 2) You apply a net force to a mass causing it to speed up You do work by applying a net force over a displacement. The object gains kinetic energy. The loss in energy by you is gained by the object. Note: We are ignoring energy lost due to friction (heat) or sound.

Energy changes/conservation of energy Q1) You pull your sibling with 20.0 N [E] for 10.0 m [E]. What is the work done by you? W = Fs = 20.0 N (10.0 m) = 200. J The work done by you is – 200. J

Energy changes/conservation of energy Q2) If for the previous problem, you applied the force at 30.0o N of E and 6.00 N of friction exists; what is the work done on the system? W = Fnet(s) = (Fax – Ff)(s) = [20.0N(cos 30.0o) – 6.00 N](10.0 m) = 113 J Note: the Work done by you is – 173 J (does not include friction!)

Energy changes/conservation of energy Q3) An 8.0 kg bowling ball is thrown by Fred Flinstone from rest to 30.0 m/s. What’s the change in kinetic energy of the bowling ball? ∆Ek = ½ mvf2 – ½ mvi2 = ½ (8.0 kg)(30.0 m/s)2 - 0 = 3.6 x 103 J b) What’s the work done by Fred? W = ∆Ek = -3.6x103 J

Energy changes/conservation of energy Q4a) What’s the potential energy of a 35 kg girl at the top of a waterslide if she is 50.0 m off the ground? Eg = mgh = 35 kg (9.81 m/s2) (50.0 m) = 1.7 x 104 J Q4b) What’s the girl’s speed at the bottom if no energy is lost? Total Mechanical Energy is conserved so (Eg)top + (Ek)top = (Eg)bottom + (Ek)bottom 1.7 x 104 J + 0 = 0 + ½ (35 kg) v2 v = 31 m/s

Energy changes/conservation of energy Q5) A 200.0 kg rollercoaster rolls down a track starting from rest at a height of 100.0 m. a) What is its speed at the bottom? 44.3 m/s b) What is the total mechanical energy anywhere? 1.96 x 105 J c) How fast is it at a height of 75.0 m? 22.1 m/s

Therefore, he utilized – 2.00 W of power Power is the rate of energy use: P = ∆E/ ∆t = W/ ∆t Ex1) A student pushes an object using 2.00 N over 60.0 m. If this was done in 1.00 minute, what power did he develop? P = W/t = Fd/t = 2.00 N(60.0 m)/[1.00 min *60s/min] = 2.00 W Therefore, he utilized – 2.00 W of power

power Ex2) A waterfall has water falling at 10.0 kg/s from a height of 1000. m. What power is developed in a turbine at the bottom, assuming the turbine is 100% efficient? P = E/t = mgh/t = 10.0 kg (9.81 m/s2) (1000 m)/1 s = 9.81 x 104 W Note: if the turbine was 33% efficient: Pout = e(Pin) = 0.33(9.81 x 104 W) = 3.24 x 104 W