North Berwick High School Department of Physics Higher Physics Unit 1 Our Dynamic Universe Section 2 Forces, Energy and Power Section 2 Forces, Energy and Power Note Making Make a dictionary with the meanings of any new words. Newton’s 1st Law of Motion 1. State Newton’s first law of motion Newton’s 2nd Law of Motion 1. State Newton’s second law of motion and write down the formula. 2. Copy and answer question 5 on page 13 as an example. Free Body Diagrams 1. Copy the three examples of free body diagrams. Conservation of Energy 1. State that the total energy of a closed system must be conserved and copy the example. Power 1. State the definition of Power and show how the formulae relate to each other. Section 2 Forces, Energy and Power Contents Content Statements ......................................................................................................... 1 Newton’s 1st Law of Motion ............................................................................................ 3 Newton’s 2nd Law ............................................................................................................ 3 Free Body Diagrams .......................................................................................................... 4 Resolution of a Force ........................................................................................................ 7 Force Acting Down a Plane ............................................................................................... 8 Conservation of Energy .................................................................................................... 9 Power .............................................................................................................................. 11 Problems ......................................................................................................................... 12 Solutions ......................................................................................................................... 22 Content Statements Content Notes Context a) Balanced and Forces acting in one Forces in rocket motion, jet unbalanced forces. dimension only. Analysis engine, pile driving, and of motion using Newton’s sport. First and Second Laws. Friction as a force acting in Space flight. The effects of a direction to oppose friction. motion. No reference to Analysis of skydiving and Terminal velocity. static and dynamic friction. parachuting, falling Tension as a pulling force raindrops, scuba diving, lifts exerted by a string or cable and haulage systems. on another object. Velocity-time graph of falling object when air resistance is taken into account, including changing the surface area of the falling object. Analysis of the motion of a rocket may involve a constant force on a changing mass as fuel is used up. b) Resolving a force Forces acting at an angle to Vehicles on a slope and ski into two tows, structures in the direction of movement. perpendicular equilibrium, e.g. supported components. The weight of an object on masts. a slope can be resolved into a component acting down the slope and a component acting normal to the slope. Systems of balanced forces with forces acting in two dimensions. 1 c) Work done, Work done as transfer of Investigating energy lost in potential energy, energy. bouncing balls. kinetic energy and power. Conservation of energy. Rollercoasters, objects in freefall. Energy and speeding vehicles. 2 Section 2 Forces, Energy and Power Dynamics deals with the forces causing motion and the properties of the resulting moving system. Newton’s 1st Law of Motion Newton’s 1st law of Motion states that an object will remain at rest or travel with a constant speed in a straight line (constant velocity) unless acted on by an unbalanced force. Newton’s 2nd Law Newton’s 2nd law of motion states that the acceleration of an object: varies directly as the unbalanced force applied if the mass is constant varies inversely as the mass if the unbalanced force is constant. These can be combined to give a α F m a = kF where k is a constant m kF = ma The unit of force, the newton is defined as the resultant force which will cause a mass of 1 kg to have an acceleration of 1 m s-2. Substituting in the above equation. k × 1 = 1 × 1 k = 1 Provided F is measured in newtons, the following equation applies. N F = ma m s-2 kg 3 Free Body Diagrams It is very rare for an object to have only one force acting upon it. More commonly, objects will have more than one force acting upon them. It is advisable to draw a diagram of the situation showing the direction of all forces present acting through one point (the centre of mass). These are known as free body diagrams. Examples 1. On take-off, the thrust on a rocket of mass 8000 kg is 200,000 N. Find the acceleration of the rocket. Thrust = 200 000 N Weight = mg = 8 000 × 9.8 = 78 400 N Resultant force = 200 000 – 78 400 = 121 600 N a = F m a = 121 600 8 000 a = 15.2ms-2 4 2. A woman is standing on a set of bathroom scales in a stationary lift (a normal everyday occurrence!). The reading on the scales is 500 N. When she presses the ground floor button, the lift accelerates downwards and the reading on the scales at this moment is 450 N. Find the acceleration of the lift. Force upwards = 500 N Lift is stationary, forces balanced W = F = 500 N Weight = 500 N F upwards = 450 N Lift accelerates downwards, unbalanced force acts. Fu Resultant Force = W - Force from floor = W – F = 500 – 450 W = 500 N = 50N a = Fu m a = 50 9.8 (from W =mg) a = 1 ms-2 5 3. Tension A ski tow pulls 2 skiers who are connected by a thin nylon rope along a frictionless surface. The tow uses a force of 70 N and the skiers have masses of 60 kg and 80 kg. Find a) the acceleration of the system b) the tension in the rope. 70N 60kg 80kg a) Total mass, m = 140 kg a = Fu m a = 70 140 a = 0.5 ms-2 b) The tension in the rope provides the necessary force to accelerate the 60 kg skier. Consider the 60 kg skier alone. Tension, T = ma = 60 × 0.5 = 30 N 6 Resolution of a Force In the previous section, a vector was split into horizontal and vertical components. This can obviously apply to a force. Fv = F sin θ θ Fh = F cos θ Example A man pushes a lawnmower of mass 100 kg with a force of 200 N acting at 30° to the horizontal. If there is a frictional force of 100 N between the lawnmower and the ground, what is the acceleration of the lawnmower along the ground? Fh = F cos θ = 200 cos 30° 30° = 173.2 N Resultant Fh = 173.2 - Friction = 173.2 - 100 = 73.2 N Friction = 100 N Fh a = Fu m = 73.2 100 = 0.732 ms-2 7 Force Acting Down a Plane If an object is placed on a slope then its weight acts vertically downwards. A certain component of this force will act down the slope. The weight can be split into two components at right angles to each other. Component of weight down slope = mg sinθ Component perpendicular to slope = mg cosθ Example A wooden block of mass 2 kg is placed on a slope at 30° to the horizontal as shown. A frictional force of 4 N acts up the slope. The block slides down the slope, from rest, for a distance of 3 m. Determine the speed of the block at the bottom of the slope. 3 m 2 kg Friction = 4 N 30° Component of weight acting down slope = mg sin 30° = 2 × 9.8 × 0.5 = 9.8 N Resultant force down slope = 9.8 - friction = 9.8 - 4 = 5.8 N 8 a = Fu m = 5.8 2 = 2.9 ms-2 v2 = u2 + 2as = 0 + (2 x 2.9 x 3) = 17.4 v = 4.2 ms-1 Conservation of Energy The total energy of a closed system must be conserved, although the energy may change its form. The equations for calculating kinetic energy Ek, gravitational potential energy Ep and work done are given below. 1 2 Ep = mgh E = mv work done = force × displacement k 2 Energy and work are measured in joules J. 9 Example A trolley of mass 1 kg is released down a slope from a height of 0.3 m. If its speed at the bottom is found to be 2 ms-1, Find a) the energy difference between the Ep at top and Ek at the bottom. b) the work done by friction c) the force of friction on the trolley 2 m 1 kg 0.3 m a) Ep at top = mgh = 1 × 9.8 × 0.3 = 2.94 J 2 Ek at bottom = mv = × 1 × 4 = 2 J Energy difference = 0.94 J 1 b) Work done by friction = energy 2difference (due to heat, sound) = 0.94 J c) Work done = Force of friction × d 0.94 = F x 2 F = 0.94 2 Force of friction = 0.47 N 10 Power Power is the rate of transformation of energy from one form to another. energy work done F displacement P = = = = F average velocity time time t Power is measured in watts W. 11 Problems Balanced and unbalanced forces 1. State Newton’s 1st Law of Motion. 2. A lift of mass 500 kg travels upwards at a constant speed. Calculate the tension in the cable that pulls the lift upwards. 3. (a) A fully loaded oil tanker has a mass of 2·0 × 10 8 kg. As the speed of the tanker increases from rest to a steady maximum speed of 8.0 m s 1 the force from the propellers remains constant at 3.0 × 10 6 N. 6 2·0 × 108 kg 3.0 × 10 N Force from propellers (i) Calculate the acceleration of the tanker just as it starts from rest. (ii) What is the size of the force of friction acting on the tanker when it is travelling at the steady speed of 8.0 m s 1? (b) When its engines are stopped, the tanker takes 50 minutes to come to rest from a speed of 8.0 m s 1.