ENERGY Energy is ‘something’ that which can be converted into work. When something has energy, it is able to perform work or, in a general sense, to change some aspect of the physical world.

KINETIC ENERGY: K, energy possessed by a body by virtue of its motion. Units: Joules (J) GRAVITATIONAL POTENTIAL ENERGY: Ug, energy possessed by a system by virtue of position or condition. Units: Joules (J) SPRING POTENTIAL ENERGY: PE, energy possessed by a system by virtue of compression or expansion of a spring. Units: Joules (J)

WORK In order for work to be done, three things are necessary: There must be an applied force. The force must act through a certain distance, called the displacement. The force must have a component along the displacement.

The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement: Work = Fd cos(q) F q Fcos(q) d Work is a scalar quantity Work in Nm = kg m/s2 x m = kg m2/s2 1 Nm = 1 Joule (J)

As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion. Work done by forces that oppose the direction of motion, such as friction, will be negative.

Work You are holding a book in the elevated position. What can be said about the Work that you are doing on the book? Positive Negative Work is done, but whether it is positive or negative can’t be determined No Work is done None of these can be determined

Centripetal forces do no work, as they are always perpendicular to the direction of motion.

When you were younger, your uncle grabbed onto your hands and swung you around and around in a circle at a constant speed. How much work is done on you when you have traveled only half of the circular path? W = Fapp (2r) W = Fapp (2πr) W = Fapp (πr) W = 0

If the force acting on an object varies in magnitude and/or direction during the object’s displacement, graphical analysis can be used to determine the work done. F is plotted on the y-axis and the distance through which the object moves is plotted on the x-axis. The work done is represented by the area under the curve.

Work Done by a Constant Force Work done by forces that oppose the direction of motion, such as friction, will be negative. f v d

You are pulling a 50kg box by a rope at 60° with 600N across a floor that has 200N of friction for 20m. How much Work is the Force of Tension doing on the box? 12000 J 10400 J 6000 J 0 J None of these

You are pulling a 50kg box by a rope at 60° with 600N across a floor that has 200N of friction for 20m. How much Work is friction doing on the box? 4000 J – 4000 J 3064 J – 3064 J 0 J None of these

You are pulling a 50kg box by a rope at 60° with 600N across a floor that has 200N of friction for 20m. How much Work does Fg do on the box? 9800 J 7507 J 1000 J 0 J None of these

You are pulling a 50kg box by a rope at 60° with 600N across a floor that has 200N of friction for 20m. What is the net (total) Work done on the box? 2000 J 8000 J 10000 J 14000 J None of these

Work When you are moving your huge entertainment center, it starts sliding down the ramp! You jump in and try to stop it by pushing on it before it hits the ground. While it is still sliding (there is friction), how many forces are doing work on the box? No forces 1 force 2 forces 3 forces 4 forces Can’t be determined

Work When you are moving your huge entertainment center, it starts sliding down the ramp! You jump in and try to stop it by pushing on it before it hits the ground. While it is still sliding (there is friction), how many forces are doing negative work on the box? No forces 1 force 2 forces 3 forces 4 forces Can’t be determined

Work When you are moving your huge entertainment center, it starts sliding down the ramp! You jump in and try to stop it by pushing on it before it hits the ground. While it is still sliding (there is friction), how many forces are doing positive work on the box? No forces 1 force 2 forces 3 forces 4 forces Can’t be determined

Forces acting: Ff F Fg and FN FN does NO work. A push of 200 N moves a 100 N block up a 30inclined plane at a constant speed. The coefficient of kinetic friction is 0.25 and the length of the plane is 12 m. a. Find the work done by each force acting on the block. F2 = 200 N Fg = 100 N θ = 30˚ μ = 0.25 d = 12 m Forces acting: Ff F Fg and FN FN does NO work. Assumes first law as v is constant???

WFg = Fg d = -Fgx d = - Fgsin30˚d = - 100 sin 30˚ (12) = - 600 J Fgy W F = F d = 200 (12) = 2400 J Ff = μ FN = μ Fgcos 30˚ = 0.25 (100) cos 30˚ = 21.6 N WFf = -Ff d = - 21.6 (12) = - 259.2 J FN F WFg = Fg d = -Fgx d = - Fgsin30˚d = - 100 sin 30˚ (12) = - 600 J Fgy θ Ff Fg Fgx

Net work: ΣW = 2400 - 259.2 - 600 = 1540.8 J FN F The resultant force: b. Show that the net work done by these forces is the same as the work of the resultant force. Net work: ΣW = 2400 - 259.2 - 600 = 1540.8 J FN F The resultant force: ΣFx = F - Ff - Fgx = 200 - 21.6 - 50 = 128.4 N W = ΣFx . d = 128.4 (12) = 1540.8 J Fgy θ Ff Fg Fgx