Work, Energy and Power Cheat Sheet PDF - Edexcel FM1

Work, Energy and Power Cheat Sheet...

Work, Energy and Power Cheat Sheet Edexcel FM1 In this chapter, we will learn to solve problems regarding the motion of a particle by considering its energy. This Power chapter is split up into three parts: In the first, we will discuss kinetic and gravitational potential energies and learn Power is defined as the rate of doing work. When considering problems involving the motion of a moving vehicle: Conservation of energy about the idea of “work done”. We will then look at how we can use the conservation of energy and work-energy The principle of conservation of energy states that: principle to solve more complicated problems. Finally, we will learn to calculate the power produced by an engine  You need to model the driving force of the vehicle, when drawing force diagrams. and use this to solve problems regarding the motion of a moving vehicle.  When no external forces (besides gravity) do work on a particle during its motion, the total energy possessed by the particle remains constant. You will also need to make use of the following relationship between the power produced by the engine and the Kinetic and potential energy driving force: At any given time, a particle possesses only two types of energy: kinetic and potential (also called gravitational This idea is useful when tackling problems where no external forces act on a particle. potential energy, or G.P.E for short)  𝑃𝑃 = 𝐹𝐹𝐹𝐹, where 𝑃𝑃 is the power of the engine, 𝐹𝐹 is the driving force and 𝑣𝑣 is the speed of the vehicle. 1 Example 3: A smooth plane is inclined at 30° to the horizontal. A particle of mass 0.5 kg slides down a line of  The kinetic energy (𝐾𝐾. 𝐸𝐸.) of a particle is given by 𝐾𝐾. 𝐸𝐸 = 𝑚𝑚𝑣𝑣 2 , where 𝑚𝑚 is its mass and 𝑣𝑣 its speed. greatest slope of the plane. The particle starts from rest at point 𝐴𝐴 and passes point 𝐵𝐵 with a speed 6 ms −1. Find Power is measured in watts (W). Questions will often state the power of a vehicle in kilowatts (kW). In such 2 the distance 𝐴𝐴𝐴𝐴. cases you need to make sure you use watts in your working. Remember that 1 kW = 1000 W  The potential energy (𝑃𝑃. 𝐸𝐸. ) of a particle is given by 𝑃𝑃. 𝐸𝐸. = 𝑚𝑚𝑚𝑚ℎ, where ℎ is the height of the particle above a fixed zero level. Example 5: A van of mass 1250 kg is travelling along a horizontal road. The van’s engine is working at 24 kW. We start again with a detailed diagram. Taking B to be the The constant resistance to motion has magnitude 600 N. Calculate: zero P.E level. a) the acceleration of the van when it is travelling at 6 ms−1. Since both of the above quantities represent energy, they are both measured in Joules (J). b) the maximum speed of the van. Remember that when calculating the potential energy of a particle, you must choose a zero level (i.e. a fixed level At A: 1 𝐾𝐾. 𝐸𝐸 = (0.5)(0)2 = 0 𝐽𝐽 2 where 𝑃𝑃. 𝐸𝐸 = 0). This can be anywhere, but it is conventional to choose the zero level to be the lowest point in the We find the kinetic and potential energy of the particle at 𝑃𝑃. 𝐸𝐸 = 𝑚𝑚𝑚𝑚ℎ = 0.5𝑔𝑔(𝑑𝑑 sin 30) particle’s motion. Be aware that potential energy can be negative, if the particle falls below the zero level. A, then at B. 1 1 a) We start with a detailed diagram. At B: 𝐾𝐾. 𝐸𝐸 = 𝑚𝑚𝑣𝑣 2 = (0.5)(6)2 = 9 𝐽𝐽 2 2 𝑃𝑃. 𝐸𝐸 = 𝑚𝑚𝑚𝑚ℎ = (0.5)(𝑔𝑔)(0) = 0 Work done Using the conservation of energy, we can use the fact that Total energy at 𝐴𝐴 = 0.5𝑔𝑔𝑔𝑔 sin 30 = 0.25𝑔𝑔𝑔𝑔 24000 = 𝐹𝐹𝐹𝐹 the total energy at A = total energy at B, since no external Total energy at 𝐵𝐵 = 9 Work done is simply just the transfer of energy from one object to another. We say that work is done on an object Use 𝑃𝑃 = 𝐹𝐹𝐹𝐹. At 𝑣𝑣 = 6, 24000 = 𝐹𝐹(6) forces act on the particle (the plane is smooth). ∴ 0.25𝑔𝑔𝑔𝑔 = 9 24000 when a force causes it to move a particular distance. To calculate the work done by a force 𝐹𝐹 on an object: 9 ∴ 𝐹𝐹 = = 4000 N Rearrange for 𝑑𝑑. 𝑑𝑑 = = 3.67 𝑚𝑚 6 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑖𝑖𝑖𝑖 0.25𝑔𝑔 𝐹𝐹 − 600 = 1250(𝑎𝑎) Use 𝐹𝐹 = 𝑚𝑚𝑚𝑚.  𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 × 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 4000 − 600 = 1250(𝑎𝑎) 3400 Solve for 𝑎𝑎. ⇒ 𝑎𝑎 = = 2.72 ms −2 Work-energy principle 1250 You may sometimes be asked to calculate the “work done against gravity”, when a particle is raised vertically. It is of course more realistic to expect a scenario where a particle is subject to an external force, whether that be 24000 = 𝐹𝐹𝐹𝐹 b) We again use 𝑃𝑃 = 𝐹𝐹𝐹𝐹, but we make 𝐹𝐹 the subject. 24000 a resistive force such as friction or perhaps an applied force acting upon the particle. For such problems, the ∴ 𝐹𝐹 = 𝑣𝑣  The work done against gravity when a particle is raised vertically is equal to its change in potential work-energy principle is very useful: 𝐹𝐹 − 600 = 1250(𝑎𝑎) energy. In other words, work done against gravity = 𝒎𝒎𝒎𝒎𝒎𝒎, where ℎ is the height raised. Use 𝐹𝐹 = 𝑚𝑚𝑚𝑚. 24000  The change in the total energy of the particle is equal to the work done on the particle. − 600 = 1250(𝑎𝑎) 𝑣𝑣 24000 Example 1: A sledge is pulled 15 𝑚𝑚 across a smooth sheet of ice by a force of magnitude 27 𝑁𝑁. The force is But note that at the maximum speed, 𝑎𝑎 = 0. − 600 = 0 When approaching questions where an external force acts on a particle, the general procedure is to: 𝑣𝑣 inclined at 25° to the horizontal. By modelling the sledge as a particle, calculate the work done by the force. 24000 = 600𝑣𝑣 1. Find the total energy of the particle at the beginning and end of the described motion. Rearrange for 𝑣𝑣. ∴ 𝑣𝑣 = 24000 = 40 ms −1 2. Find the change in energy, by subtracting the final energy from the initial energy. 600 3. Find the work done by the external force on the particle. We start with a detailed diagram. 4. Equate this to the change in energy and solve for any unknown constants. Example 6: A car of mass 2600 kg is travelling in a straight line. At the instant when the speed of the car is 𝑣𝑣 ms−1 , the total resistances to motion are modelled as a variable force of magnitude (800 + 5𝑣𝑣 2 ) N. The car has a cruise The box moves across the ice, so the work done is Example 4: A box of mass 2 kg is projected with speed 6 ms −1 up a line of greatest slope of a rough plane inclined control feature which adjusts the power generated by the engine to maintain a constant speed of 18 ms−1. Find 1 going to be equal to the horizontal component of 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = (27 cos 25) × 12 = 293.6 𝐽𝐽 at 30° to the horizontal. The coefficient of friction between the box and the plane is. the power generated by the engine when: 3 the force multiplied by the distance travelled. a) Use the work-energy principle to calculate the distance the box travels up the plane before coming to rest. a) the car is travelling on a horizontal road. b) Suggest why in practice the box may not travel as far as the distance you calculated. b) the car is travelling up a road that is inclined at an angle 4° to the horizontal. Example 2: A package of mass 2𝑘𝑘𝑘𝑘 is pulled at a constant speed up a rough plane which is inclined at 30° to the horizontal. The coefficient of friction between the package and the surface is 0.35. The package is pulled 12𝑚𝑚 up a) We start again with a detailed diagram. Take the starting point to be the zero P.E level. Let 𝑑𝑑 represent the distance a line of greatest slope of the plane. Calculate: a) the work done against gravity. b) the work done against friction. travelled up the plane till the point of instantaneous rest. a) We start with a detailed diagram. 1 At start: 𝐾𝐾. 𝐸𝐸 = (2)(6)2 = 36 𝐽𝐽 2 𝑃𝑃. 𝐸𝐸 = 𝑚𝑚𝑚𝑚ℎ = 2𝑔𝑔(0) = 0 Use 𝐹𝐹 = 𝑚𝑚𝑚𝑚 𝐹𝐹 − (800 + 5𝑣𝑣 2 ) = 2600(𝑎𝑎) We find the kinetic and potential energy of the particle at A, then at B. 1 1 We are told 𝑣𝑣 = 18 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐), and thus 𝑎𝑎 = 0. Substitute in 𝐹𝐹 − (2420) = 2600(0) a) We start with a detailed diagram. At end: 𝐾𝐾. 𝐸𝐸 = 𝑚𝑚𝑣𝑣 2 = (2)(0)2 = 0 𝐽𝐽 these values. 𝐹𝐹 = 2420 2 2 𝑃𝑃. 𝐸𝐸 = 𝑚𝑚𝑚𝑚ℎ = (2)(𝑔𝑔)(𝑑𝑑 sin 30) = 𝑔𝑔𝑔𝑔 𝑃𝑃 = 𝐹𝐹𝐹𝐹 ∆𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 36 − 𝑔𝑔𝑔𝑔 Use 𝑃𝑃 = 𝐹𝐹𝐹𝐹. Next, find the change in energy and the work done by the 𝑃𝑃 = (2420)(18) = 43 560 W = 43.6 kW (3 𝑠𝑠. 𝑓𝑓. ) frictional force. Remember that Work done by resistive force = 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 × 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = (2)(𝑔𝑔)(12 sin 30) change in energy = energy before – energy after = 𝐹𝐹 × 𝑑𝑑 Work done against gravity = 𝑚𝑚𝑚𝑚ℎ We also use 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚 = 𝜇𝜇𝜇𝜇 since friction is limiting. 1 = 𝜇𝜇𝜇𝜇 × 𝑑𝑑 = 𝑅𝑅𝑅𝑅 = 12𝑔𝑔 𝑁𝑁 3 b) We draw another detailed diagram. 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑏𝑏𝑏𝑏 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝐹𝐹 × 𝑑𝑑 𝑅𝑅 − 2𝑔𝑔 sin 30 = 0 b) Use work done = force × distance Find R by resolving perpendicular to the plane: ∴ 𝑅𝑅 = 2𝑔𝑔 sin 30 = 𝑔𝑔 = (0.35)𝑅𝑅(12) 1 So, the work done by the frictional force becomes: ∴ work done = 𝑔𝑔𝑔𝑔 𝑅𝑅 − 2𝑔𝑔 cos 30 = 0 3 We resolve perpendicular to the plane to find Use 𝐹𝐹 = 𝑚𝑚𝑚𝑚 up the slope. 𝐹𝐹 − 2600𝑔𝑔 sin 4 − (800 + 5𝑣𝑣 2 ) = 2600𝑎𝑎 ∴ 𝑅𝑅 = 2𝑔𝑔 cos 30 ∆𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑏𝑏𝑏𝑏 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑅𝑅. Use the work-energy principle. 1 ⇒ 36 − 𝑔𝑔𝑔𝑔 = 𝑔𝑔𝑔𝑔 We are told 𝑣𝑣 = 18 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐), and thus 𝑎𝑎 = 0. Substitute in 𝐹𝐹 − 10400𝑔𝑔 sin 4 − 2420 = 0 3 these values. ∴ 𝐹𝐹 = 9530 N 1 36 Calculate work done. 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = (0.35)(2𝑔𝑔 cos 30)(12) = 71.3 𝐽𝐽 Solve for 𝑑𝑑. 𝑑𝑑 𝑔𝑔 + = 36 ⇒ 𝑑𝑑 = 1 = 3.55𝑚𝑚 𝑃𝑃 = 𝐹𝐹𝐹𝐹 3 𝑔𝑔 + Use 𝑃𝑃 = 𝐹𝐹𝐹𝐹. 3 𝑃𝑃 = (9530)(18) = 171 532.. = 170 kW (3 𝑠𝑠. 𝑓𝑓. ) In reality air resistance is likely to oppose the motion of b) This is a common answer to such questions. the particle causing it to travel a shorter distance than the one we found. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc