Work and energy

Using graphs to summarise motion 22/05/18 © Class Leading Ltd. 2013. Permission granted for non-commercial educational use provided that this copyright notice is included. Using graphs to summarise motion D Distances measured in one direction are ________, and in the other are _______ The _______ of an object is its ______ and the ________ of motion. The velocity of an object moving in a straight line is ________ if it is moving in one direction and ________ if it is moving in the opposite direction. Sketch a velocity-time graph for an object that is: stationary moving in a straight line with constant speed moving in a straight line with steadily increasing Sketch a distance-time graph for an object that is: moving at constant speed moving with increasing speed Using the graph to the right which is moving faster, A or B? Calculate the speed of B C v t b a c speed negative direction d t c a b positive velocity negative A/B positive time/s distance /m A B 10 20 100 200 A – steeper slope / gradient Speed = gradient = (100-0) = 100 = 5 m/s (20-0) 20

Doing work: When you push or pull something and make it move, you do work.

Energy and work Whenever an object moves energy is used This also means that work has been done E.g you open a door You have used force to open the door You have used up energy to open the door Therefore you have done work The energy has been transferred to opening the door

When an object is moved by a force, we say work is done on the object by the force The work done on the object is equal to the energy transferred to the object E.g. the energy you use to open the door is transferred to the door (door moves) If the energy you used up to open the door was 20J then the work done was 20J Where does the energy come from? When you do work against friction, most of the energy gets lost as heat and sometimes as sound

work done (in joules) = force (in newtons)  distance (in metres) How much work? Imagine pushing a car along a road. You must apply a force . . . . . . and you push the car a certain distance. work done (in joules) = force (in newtons)  distance (in metres)

LQ: Can I calculate the energy used in different situations? Work done, GPE and KE LQ: Can I calculate the energy used in different situations?

An example I push my car 30 m. I push with a force of 400 N. work done = force  distance = 400 N  30 m = 12 000 J

Lifting something If you lift something up, you must exert enough force to balance the force of gravity. The upward force is equal to the weight of the object.

work done = force  distance force = weight of the object lifted Lifting something If you lift something through a certain distance work done = force  distance where force = weight of the object lifted

An example A woman lifts up her baby from the floor. The baby weighs 100 N. The woman lifts the baby up by 2 m. work done = force  distance = weight  distance = 100 N  2 m = 200 J

amount of work done = amount of energy transferred Work done = energy transferred When someone does work, energy is transferred: from the person who exerts the force to other places. amount of work done = amount of energy transferred

Let’s look first at lifting distance lifted When you lift something up, you do work. work done = force  distance = weight  distance lifted The gravitational potential energy of the object you lift has increased: change in gravitational potential energy = weight  height gain

work done = force  distance moved in the direction of the force Different paths? What if you lift something, but not straight up? work done = force  distance moved in the direction of the force change in gravitational potential energy = weight  vertical height gain

Hannah pushes a book 5m along the table with a force of 5N Hannah pushes a book 5m along the table with a force of 5N. She gets tired and decides to call it a day. How much work did she do? Courtney lifts a laptop 2m into the air with a force of 10N. How much work does she do? What type of energy did the laptop gain? Tom does 200J of work by pushing a wheelbarrow with a force of 50N. How far did he push it? What type of energy did the wheelbarrow gain? Dan cuddles his cat and lifts it 1.5m in the air. If he did 75J of work how much force did he use? Simon drives his car 1000m. If the engine was producing a driving force of 2000N how much work did the car do? 25J 20J, GPE 4m, KE 50N 2MJ

Some example questions… How much gravitational potential energy have the following objects gained?: A brick that weighs 10N lifted to the top of a house (10m), A 1,000kg car lifted by a ramp up to a height of 2m, A 70kg person lifted up 50cm by a friend. How much GPE have the following objects lost?: A 2N football dropping out of the air after being kicked up 30m, A 0.5N egg falling 10m out of a bird nest, A 1,000kg car falling off its 200cm ramp. 100J 20KJ 350J 60J 5J 20KJ

work done = change in kinetic energy of the cart Work done = energy transferred What happens if you push something horizontally? Imagine pushing a cart with very well-oiled wheels (no friction). The speed of the cart will increase while the force acts. work done = change in kinetic energy of the cart

kinetic energy = ½ mass  (speed)2 Calculating kinetic energy The kinetic energy of a moving object depends on: its mass (how big it is) its speed (how fast it is moving). The equation for calculating the kinetic energy of a moving object is: kinetic energy = ½ mass  (speed)2

the amount of work you do = the amount of energy transferred What if friction cannot be ignored? Think about pushing a car. As you push it, you transfer energy from your body to: the car – increasing its kinetic energy parts of the car (and the surroundings) – which heat up because of friction. the amount of work you do = the amount of energy transferred

Now let’s look at an example that brings several of these ideas together. . . . . . pushing a car up a slope.

If you push your car uphill, you do work: work done = force  distance Pushing a car uphill If you push your car uphill, you do work: work done = force  distance You increase the gravitational potential energy of your car: change in gravitational potential energy = weight  vertical height gain Are these the same? If not, why not?

The calculation work done in pushing = force  distance = 450 N  30 m = 13 500 J gain in gravitational potential energy = weight  vertical height gain = 900 N  1 m = 9000 J

So why the difference? Some of the energy is transferred to the surroundings – heating them up. Only part of it goes to increasing the gravitational potential energy of the car. If no energy is lost as heat you can assume that kinetic energy = gravitational potential energy (for falling objects only)

A diver who has a mass of 50 kg dives off a diving board 3 A diver who has a mass of 50 kg dives off a diving board 3.0 metres above the water level. What is her kinetic energy when she reaches the water? A diver who has a mass of 50 kg dives off a diving board 3.0 metres above the water level. What is her kinetic energy when she reaches the water?

Kinetic energy gained = gravitational potential energy lost Kinetic energy gained = weight × height You must calculate her weight to use in this equation Weight = mass × gravitational field strength Weight = 50 kg × 10 N / kg Weight = 500 N Kinetic energy gained = 500 N × 3 m Kinetic energy gained = 1500 J

How fast is the diver moving when she reaches the water? Ke = 0.5mv2

Put her kinetic energy (the 1500J you have calculated) into the kinetic energy equation together with her mass Kinetic energy = 1/2 × mass × speed2 1500 J = ½ × 50 × speed2 = 25 × speed2 So speed2 = 1500/25 = 60 (This is not the answer yet! It is speed2!) So her speed = square root of 60 = 7.7 m/s