Think back to GCSE… Define the following terms and see if you can make any connections between them. Energy Work Power Time Distance Force Velocity

Work and Power define work done by a force (b) define the joule (c) calculate the work done by a force using W = Fs and W = Fs cos θ (d) define the term power as the rate of energy used. (e) calculate the power by using the P = E/t and P = Fv

Energy can be transferred from one object to another by means of Work done Heat transfer Also electricity, sound waves and electromagnetic radiation transfer energy.

We say that work is done by a force when the object concerned moves in the direction of the force, and the force thereby transfers energy from one object to another. What sort of energy is gained by raising a mass? What is the force acting against the movement? We say that the lifting force is doing work against gravity. If the object is raised at a steady speed, it is in equilibrium then the lifting force will just balance weight. This of course ignores any air resistance etc. What sort of energy is gained by raising a mass (board rubber etc)? (GPE) What is the force acting against the movement? (mg) We say that the lifting force is doing work against gravity. Some students may feel that it takes a larger force than mg to raise the object; however, if the object is raised at a steady speed, it is in equilibrium and the lifting force will just balance weight. This of course ignores any air resistance etc What is g? (You are looking for gravitational field strength here, not acceleration due to gravity; you may need to draw this point out.) What two factors will determine how much energy you lose (and hence the energy transferred to the object)? (The size of the force and the distance h moved in the direction of the force.) Can you state the equation which gives the energy gained? (mgh) Hence how much energy is required to lift object height h? (mgh) This suggests that energy transferred = force ´ distance moved in direction of force. This is known as the work done by the force. In fact, we have used the fact that students are familiar with the equation GPE = mgh to lead them to the more general equation for work done. You should now point out that mgh is simply a particular case of work done. You can now push the board rubber or thunderbird across the lab bench to show work being done against a frictional force. In this case, the energy transferred ends up as heat. What is g? Gravitational field strength

Can you state the equation which gives the energy gained? What two factors will determine how much energy you lose (and hence the energy transferred to the object)? Can you state the equation which gives the energy gained? Hence how much energy is required to lift object height h? This suggests that energy transferred = force x distance moved in direction of force. This is known as the work done by the force. In fact, we have used GPE = mgh to lead us to the more general equation for work done. GPE = mgh is simply a particular case of work done.

What force is work being done against when a block is being dragged across a frictional surface? Note: work is energy changed from one form to another – it’s not necessarily the total energy. Work being done against a frictional force. In this case, the energy transferred ends up as heat.

𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒=𝑓𝑜𝑟𝑐𝑒×𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 Quick Question An object is moved a distance of 10cm. If the force applied is 30kN what is the work done? Note: the equation assumes the direction of the force is the same as the direction of movement. Note: the equation also gives us the definition of the joule (J): ‘one joule is the work done when 1 newton moves an object through a distance of 1 metre’.

What’s the issue here?

Standard Q.1 and 2 on Pg.159 Q1. W = Fs = 203 x 2.81 = 570.43 = 570J (3 s.f.) Q2. W = Fscosθ = 371 x 1.29 x cos13.1 = 502.269... = 502 J (3 s.f.) Challenge Q. 2 on Pg.159 Q2. W = Fscosθ = 371 x 1.29 x cos13.1 = 502.269... = 502 J (3 s.f.)

Force-Distance Graphs What does the area underneath represent? Force, F Distance, s What does the area underneath represent?

Force-Distance Graphs Force, F Extension, ΔL 𝐴𝑟𝑒𝑎= 1 2 𝐹∆𝐿

Force-Distance Graphs

How would you calculate the gain in potential energy? How could you calculate the power of this?

Energy transferred per second is the power of the objects 𝑝𝑜𝑤𝑒𝑟= 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑖𝑚𝑒 𝑝𝑜𝑤𝑒𝑟= 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑡𝑖𝑚𝑒

Work done per second moving the plane forward; =𝑓𝑜𝑟𝑐𝑒×𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 Therefore; 𝑝𝑜𝑤𝑒𝑟=𝑓𝑜𝑟𝑐𝑒×𝑠𝑝𝑒𝑒𝑑

This machine is applying a force to the object to move it This machine is applying a force to the object to move it. This force is causing the object to move through a distance, s. 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒, 𝑊=𝐹𝑠

Providing the object is moving at a constant velocity v due to the force; 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑠=𝑣𝑡 This is in a set time, t

Note: the equation assumes the direction of the force is the same as the direction of movement. You may need to calculate the component of the force in the direction of movement. The output power of the digger becomes; 𝑃𝑜𝑤𝑒𝑟= 𝑊.𝑑. 𝑡 = 𝐹𝑣𝑡 𝑡 =𝐹𝑣 𝑃 𝑂𝑈𝑇 =𝐹𝑣

Standard Q.3 and 4 on Pg.159 Q3. P = Fv, F = P/v = 60100/34.7 = 1731.98... = 1730 N (3 s.f.) Q4. On the right: Challenge Q. 4 and 5 on Pg.159 Q4. On the right: Q5. Look only at the horizontal component of the force: P = Fvcosθ = 83.1 x 2.99 x cos15.2 = 239.776... = 240 W (3 s.f.)

vector metre scalar joule scalar watt

Prep Homework Evidence that you know and understand: The meaning of and equation for efficiency Sankey diagrams I want to see diagrams, calculations etc. If you’re stuck – look back at GCSE notes!