WorkEnergy Theorem F m d In this presentation

Work-Energy Theorem F m d In this presentation you will: n investigate quantities using the work-energy theorem Next >

Introduction Many words used in everyday conversation have very precise meanings in physics. When you think of work, you may think of going to a place of work, or working hard, or even something like a computer working. rk o W Energ y Force When you think of energy, you may think of how lively or tired you are. In physics, work and energy have very specific meanings. Next >

What is work? Consider a force (F) acting on an object while it moves a distance (d). F m d The object’s velocity will change; it will accelerate. a= F m vf vi m m Using the equations of motion, we can find the change in velocity. d vf 2 – vi 2 = 2 ad Replacing a = F m and multiplying both sides by , we get: m 2 Fd = ½mvf 2 – ½mvi 2 Next >

What is work? Fd = ½mvf 2 – ½mvi 2 m The left side of the equation is the work done on the system. vf vi m F d In physics, work has a precise definition: Work is the product of force × distance. Work = F × d So: W = ½mvf 2 – ½mvi 2 Next >

Energy W = ½mvf 2 – ½mvi 2 The right side of the equation describes the change in quantity before and after the force acts. vf vi m m F d The quantity depends on the mass and the velocity of the object. This quantity is known as energy and as it is the energy of a moving body, kinetic energy. KE = ½mv 2 So: W = KEf – KEi Next >

Work-Energy Theorem Work is equal to the change in kinetic energy. W = ΔKE vf vi m m F d In physics, the Δ symbol is used to represent a change in something. The units of work and energy are the joule (J), named after the physicist James Joule, who discovered the relationship. 1 J = 1 Nm (Fd) = 1 kgm 2/s 2 (½mv 2) Next >

Calculating Work is calculated using the equation: W = Fd F m d A force of 10 N is used to move an object over a distance of 10 m. How much work is done? W = Fd = 10 × 10 = 100 J Work = 100 J Next >

Question 1 How much work is done when a force of 25 N is used to move an object a distance of 5 m? Give your answer as a number in J. Next >

Question 1 How much work is done when a force of 25 N is used to move an object a distance of 5 m? Give your answer as a number in J. W = F × d = 25 × 5 = 125 J Work = 125 J 125 (J) Next >

Friction Force In the real world, friction tends to act on a moving body to slow it down. F m Ff d Now we have a friction force (Ff) acting in the opposite direction. Ff acts in the opposite direction, so it does negative work. W = (F – Ff)d m Ff d If the friction force Ff was the only force acting on the object when it is in motion, the object would slow down, and its kinetic energy would be reduced. W = (0 - Ff)d = - Ff d Next >

Question 2 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. How much work is done on the object? Give your answer as a number in J. Next >

Question 2 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. How much work is done on the object? Give your answer as a number in J. W = -Ff × d = -10 × 10 = -100 J Work = -100 J -100 (J) Next >

Question 3 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. What is its change in kinetic energy? Give your answer as a number in J. Next >

Question 3 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. What is its change in kinetic energy? Give your answer as a number in J. W = -Ff × d = -10 × 10 = -100 J Work = ΔKE = -100 J -100 (J) Next >

Question 4 A force of 100 N is used to move an object over a distance of 10 m. A friction force of 10 N is acting on the object to slow it down. What is the work done? Give your answer as a number in J. Next >

Question 4 A force of 100 N is used to move an object over a distance of 10 m. A friction force of 10 N is acting on the object to slow it down. What is the work done? Give your answer as a number in J. W = (F-Ff) × d = (100 – 10) × 10 = 900 J Work = 900 J 900 (J) Next >

Question 5 A moving object has a kinetic energy of 100 J. A force of 30 N is applied to the object in the same direction as its travel over a distance of 10 m. What is the total kinetic energy after 10 m? Give your answer as a number in J. Next >

Question 5 A moving object has a kinetic energy of 100 J. A force of 30 N is applied to the object in the same direction as its travel over a distance of 10 m. What is the total kinetic energy after 10 m? Give your answer as a number in J. W = F × d = 30 × 10 = 300 J Work = ΔKE = 300 J Total KE = 300 + 100 = 400 J 400 (J) Next >

Force at an Angle To calculate force applied at an angle, the force is split into x and y components. F m d Only the component of the force that acts in the direction of the displacement is used. Fx θ Fy F Fx = F cosθ The Fy component acts at right angles to the direction of displacement so no work is done. So, W = Fd cosθ Next >

Force at an Angle For example, if the force of 10 N is acting at an angle of 30° over a distance of 10 m: F m d W = Fd cosθ Fx θ Fy W = 10 × cos 30° J F W = 86. 6 J Next >

Summary In this presentation you have seen: n work is an energy transfer calculated by the product of a force and distance W=F×d n the kinetic energy of a moving body is related to its mass and velocity KE = ½mv 2 n the work-energy theorem relates work and energy W = ΔKE End