#### Transcript Work=Force x Distance Power = Work/Time

Work Essential Question: What is work (in physics)? Wake Up Warm Up In a physical sense, what do you think is more work? a)Holding a 100 kg weight above your head. b)Pushing really hard against a wall to the point of exhaustion c)Lifting a 10 kg object above your head from the floor or Pushing a 50kg box across the floor at a constant velocity d)Studying all night long for a physics test Let’s create a definition The only case that involves any work is the third case. Why? What do you think is the definition of work? Work is a force action over some distance to cause a change in energy. The amount of work done depends on two things: a. The amount of force exerted and b. The distance over which the force is applied. There are two factors to keep in mind when deciding when work is being done: something has to move and the motion must be in the direction of the applied force. Work can be calculated by using the following formula: Work=force x distance Units of Work Work = Force x Distance The unit of force is Newtons (N) The unit of distance is meters The unit of work is newton-meters One Newton-meter is equal to one joule So, the unit of work is a joule (J) Work is done on the books when they are being lifted, but no work is done on them when they are being held or carried horizontally. In which of the following cases was work done: Someone applies a force to a wall and becomes exhausted. 2. A book falls off a table and free falls to the ground. 3. A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. 4. A rocket accelerates through space. 1. Let’s Practice Work = Force x Distance Calculate: If a man pushes a concrete block 10 meters with a force of 20 N, how much work has he done? Work = Force x Distance Calculate: If a man pushes a concrete block 10 meters with a force of 20 N, how much work has he done? (W = 20N x 10m =200 joules) Power and Work Power is the rate at which work is done. Force=Mass x Acceleration Work=Force x Distance Power = Work/Time The unit of power is the watt. Practice Questions 1. Two students, Ben and Bonnie, are in the weightlifting room. Bonnie lifts the 50 kg barbell over her head (approximately 60 m) 10 times in one minute; Ben lifts the 50 kg barbell the same distance over his head 10 times in 10 seconds. • Which student does the most work? • Which student delivers the most power? • Explain your answers. Practice Questions 2. How much power will it take to move a 10 kg mass at an acceleration of 2 m/s/s a distance of 10 meters in 5 seconds? This problem requires you to use the formulas for force, work, and power all in the correct order. Force=Mass x Acceleration Work=Force x Distance Power = Work/Time 2. How much power will it take to move a 10 kg mass at an acceleration of 2 m/s/s a distance of 10 meters in 5 seconds? This problem requires you to use the formulas for force, work, and power all in the correct order. Force=Mass x Acceleration Force=10 x 2 Force=20 N Work=Force x Distance Work = 20 x 10 Work = 200 Joules Power = Work/Time Power = 200/5 Power = 40 watts What if the force is at an angle to the displacement? where F is the force, d is the displacement, and the angle (theta) is defined as the angle between the force and the displacement vector Force Scenarios Scenario A: A force acts rightward upon an object as it is displaced rightward. Scenario B: A force acts leftward upon an object which is displaced rightward Scenario C: A force acts upward on an object as it is displaced rightward. The least steep incline (30-degree incline angle) will require the least amount of force while the most steep incline will require the greatest amount of force. Yet, force is not the only variable affecting the amount of work done by the car in ascending to a certain elevation. Another variable is the displacement which is caused by this force. A look at the animation above reveals that the least steep incline would correspond to the largest displacement and the most steep incline would correspond to the smallest displacement. Free Body Diagrams Diagram A Answer: W = (100 N) * (5 m)* cos(0 degrees) = 500 J The force and the displacement are given in the problem statement. It is said (or shown or implied) that the force and the displacement are both rightward. Since F and d are in the same direction,the angle is 0 degrees. Diagram B Answer: W = (100 N) * (5 m) * cos(30 degrees) = 433 J The force and the displacement are given in theproblem statement. It is said that the displacement is rightward. It is shown that the force is 30 degrees above the horizontal. Thus, the angle between F and d is 30 degrees. Diagram C Answer: W = (147 N) * (5 m) * cos(0 degrees) = 735 J The displacement is given in the problem statement. The applied force must be 147 N since the 15-kg mass (Fgrav=147 N) is lifted at constant speed. Since F and d are in the same direction, the angle is 0 degrees. Example: Force at an angle A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. How much work is done by the applied force? W = F * d * cos(θ) W = (50 N) * (3 m) * cos (30o) = 129.9 J Example force applied vertically How much work is done by an applied force to lift a 15Newton block 3.0 meters vertically at a constant speed? To lift a 15-Newton block at constant speed, 15-N of force must be applied to it (Newton's laws). Thus, W = (15 N) * (3 m) * cos (0 degrees) = 45 Joules Example 2: Force at an angle Calculate the work done by a 2.0-N force (directed at a 30° angle to the vertical) to move a 500 gram box a horizontal distance of 400 cm across a rough floor at a constant speed of 0.5 m/s. (HINT: Be cautious with the units.) Here is a good example of the importance of understanding the angle between F and d. In this problem, the d is horizontal and the F is at a 60-degree angle to the horizontal. Thus, theta is 60 degrees. W = (2.0 N) * (4.00 m) * cos (60 degrees) = 4.0 J Bibliography Credit: MFA Resources dropbox