Work happens when a force causes an object to move through a displacement. • Be careful with this definition. Some people incorrectly assume that it means a force is work. • Instead, what it really means is that a force exerted on an object can cause work to happen. • We have to be careful that the object actually has a displacement, otherwise work has not happened. • The other detail for this definition is that the force and the displacement must point in the exact same direction. • Lets look at some examples and decide if work is being done.
Example 1: I am holding a 2 kg block of cheese in my hands. I walk 12 m to the other side of the room. Explain if I did any work. • Since I am holding the cheese up against the force of gravity (Fg), the force I must be applying (FN) on it will be pointing up vertically. I moved horizontally, so the two vectors (force and displacement) are perpendicular to each other. I didn’t do any work.
Example 2: I decide to get my pet frog to do a little weight lifting (but I’m going to start him off slow!). He lifts 10 kg up from the floor, over his head, and back down to the floor. Explain if he did any work. • Well, in this case the force must be pointing up when he lifts up the weights, and at first he's moving them up, so everything seems fine so far. But wait… I said that he then brings the weights back down to the floor. Overall, the displacement of those weights is zero!
Example 3: Last winter my car got caught in a snow bank. I promise one of my friends that if he comes over to do some work for me I’ll buy him a Big Mac (with extra onions… he really likes onions). We get behind the car and push it out of the snow. Explain if we did any work. • In this situation, both of us were pushing in the same direction (parallel to each other) and the car moved in that direction. So the answer would be “Yes!” I do owe him a Big Mac for the work he did… but I’ll cut it in half and eat part of it myself since I did half the work.
Sample Problems from yesterday: 2. A person lifts a 10 kg slab a height of 2.2 metres at constant velocity. a) Calculate the work done by the person. b) Calculate the work done by the force of gravity.
Example 4: You are pulling a box with a rope at a angle from the ground (as shown in Illustration • 1). The box moves 12.7 m when you pull along the rope with a force of 76.0 N. Determine how much work you did.
3. A rope angled at above the horizontal is used to pull a sled a displacement of 30 m [forward]. The tension in the rope is 150 N. a) Calculate the work done by the tension. b) Calculate the work done by gravity. c) Calculate the work done by the normal force.
4. A block of mass 1.0 kg is on a ramp angled at . The block slides down the ramp a total displacement of 2.0 m. The coefficient of friction is 0.20. a) Calculate the work done by gravity. b) Calculate the work done by the normal force. c) Calculate the work done by friction.
Graphing Work: • Like many other concepts in physics, it is often easier to interpret information or come to conclusions if the data is presented as a graph. • This does sometimes make it harder when you are first learning physics, since we make you learn a bunch of rules for each type of graph. • You should try to see beyond the rules… if you can do this, and understand each graph for what it truly represents, you will be on the road to understanding physics.
In the case of Work, the graph that we are most interested in is Force vs. Displacement and the area underneath (or sometimes above) the line. • Imagine a really boring situation where a constant force is applied to an object that causes it to move a displacement. The graph would look something like...
If you were asked to calculate the area under the line of the graph, what would you do? • Since it is a rectangular shape, you’d probably say Area = base X height • But wait a second… the base is a displacement and the height is a force… • A = b (h) • A = F (d) • W = F d • Area under the line = Work!
This is really handy for doing calculations of work if the force is not constant. • As long as we can calculate the area under the line, we will have the work that has been done. • Watch out that you use the appropriate area formula for the graph you are looking at. • We can not just run in and use the formula W =Fd in these situations since force is not a constant.
Example 1: A rocket engine is going through a series of tests. It is ignited and data is measured as it builds up thrust. If the initial force was zero as it was switched on, and it increases constantly up to 1.6e5 N as the rocket moves 112 m along the test track, determine how much work was done. • First, we’ll graph the information.
Example 2: You press down on the gas pedal of your car and accelerate through the mob of zombies in front of you. As you move 65 m through the swarm to get out of the parkade of the shopping mall you've been hiding in, you estimate that the force of your car increases from 200 N to 780 N linearly. Determine how much work your car did to the zombies. (3.2e4 J)
A block of mass 4.5 kg is on a ramp angled at . The block is pushed up the ramp a total displacement of 6.0 m at constant velocity. The coefficient of kinetic friction is 0.25. Calculate the total work done by: a) Gravity b) Friction c) Normal force d) Applied force
Example 5: Sally pushes a 5 kg box at an angle of below the horizontal, and moves the box 25 meters in 20 seconds. The box starts from rest, and has a coefficient of friction of 0.35. a) Calculate the work done by Sally. b) Calculate the work done by the normal force. c) Calculate the work done by friction. d) Calculate the work done by gravity.