# M2: Work-energy principle and statics

LESSON 1: THE ROLLER-COASTER

Starter:

The roller-coaster track is 72m high. Assume there’s no friction. What is the speed of the car at the bottom of the track?

Revision: kinetic energy and potential energy. Units: joules.

Activity: Ex.3B Q.1,2

Activity: Ex.3C Q.1,2

LESSON 2: WORK

Starter

Suppose the roller-coaster track is 400m long, and all along it the car is slowed down by a frictional force of 30 newtons. Then what will the final speed of the car be?

Exposition: work-energy principle

KE at start + PE at start + Work put in by the person or engine − Energy dispersed through friction etc. = KE at end + PE at end

Demonstration: inclined plane and spring balance

Activity: Ex.3C Q.3, 4

Homework: Ex.3C Q.5, 6, 15, 17.

LESSON 3: WORK-ENERGY PRINCIPLE WITH SELF-POWERED OBJECTS

Starter

The roller-coaster track is 72m high and 400m long. Suppose there is no friction, but the car has an engine which applies a constant force of 30 newtons to accelerate it. What is the speed of the car at the bottom of the track?

Ex.3C Q.7-14

LESSON 4: POWER

Power is rate of doing work in joules per second. 1 joule per second = 1 watt. Power=force×speed

Why is the electricity we use measured in kilowatt hours rather than joules? How many joules in a kilowatt hour?

Guess the power (rate of working) of the following:

A compact (economy) light bulb at home

A fluorescent tube light at school

A fridge

A person walking on a flat road

A cyclist cycling on a flat road

A cyclist racing, or cycling up a steep hill

An expert weightlifter making a big lift

A car at maximum acceleration

A big truck at maximum acceleration

A Eurostar train at maximum acceleration

A compact (economy) light bulb at home – 10 to 15 watts

maths

A fluorescent tube light at school – about 40 watts

A fridge – 35 watts

A person walking on a flat road – 30 watts

A cyclist cycling on a flat road – 30 watts

A cyclist racing, or cycling up a steep hill – 200 watts

An expert weightlifter making a big lift – 5000 watts

A car at maximum acceleration – 100,000 watts

A big truck at maximum acceleration – 450,000 watts

A Eurostar train at maximum acceleration – 12,000,000 watts

Activity: Ex.3D Q.1, 2, 3, 4, 5, 7

FOLLOW-UP AND PRACTICE ON WORK, ENERGY, AND POWER

Activity: Ex.3D Q.10, 14, 15. Ex.3E Q.12, 13, 15. Review Exercise page 96 Q. 43, 44.

Issues:

DO # Calculate Work = Force ⨯ Distance in the direction of the force

DON’T # Put cos α where you should put sin α or vice versa
DO # Check by thinking: if α=0, would this PE or Work be zero?

DON’T # Forget bits when writing out the multiplications for PE, or for Work done against friction
DO # Write each element out bit by bit, e.g. Work done against friction may require multiplying together five different bits, coefficient × mass × g × cos [some angle] × distance

LESSON 5-6-7-8: WHEN RODS ARE IN EQUILIBRIUM

Note these points

If you’re asked something like “state how you have used the modelling assumption that the ladder is a uniform rod”, the answer is always: the total weight of the ladder can be assumed to act through its centre. For other “state how you have used the modelling assumption” bits, click here.

Practice on statics:

Review exercises Q.21 to 36

Summary notes

Each problem can be divided into three steps.

1. Do a diagram. Big, clear, with a ruler. Insert all your information from reading the whole question. Look out for key words like “smooth” (friction = zero), “in limiting equilibrium” or “on the point of slipping” (means friction is at its maximum value μR). If you know that a frictional force F1 = μR1, then write μR1 in on your diagram. Draw long lines of action for your forces. Mark the “α” and “not-α” angles. Resolve all your forces into horizontal and vertical components. (Not into components along the line of the rod and perpendicular: in principle there’s nothing wrong with resolving them that way, but in practice, in M2 statics problems, it is simpler and clearer just always to resolve all forces into horizontal and vertical components).

2. Write your three equations: resolve horizontally, resolve vertically, take moments. Label each equation so the reader knows what it is. Choose the point to take moments round so as to get as few unknowns as possible in that equation. (In principle, you can solve the problem even with the worst choice of point to take moments round. It’s just more complicated). Simplify your three equations (for example, by inserting “cos α” and “sin α” values if you have them) before you go on.

3. You should now have three equations in three unknowns. (There may be other letters, say W for the weight of the rod or m for its mass, which are given quantities for the problem). Now, in theory, you can forget all about your diagram. It’s just an algebra problem of solving three equations for three unknowns. Make it clear, in your working, when you are using which equation. (You may want to number them, [1], [2], [3], to help that).