- Work and power
- Work and kinetic energy relationship
- Conservation of energy
- Machines
Work and power:
Work occurs when a force is exerted over a distance, and the two have to be parallel. The equation for work is work = force x distance. The unit of measurement for work is joules or J for short. Work is responsible for power, for it measures how quickly work is done. Power is measured in watts and the equation we use to solve for power is, power = work/time.
An example problem for work is: A boy pushes a box to the right with 30 newtons for 5 meters. It takes him 10 seconds to do this. Solve for the work and the power.
Work = force x distance
Work = 30N x 10m
Work = 300Jouls
Power = work/time
Power = 300/10
Power = 30watts
Work and kinetic energy relationship:
Kinetic energy is energy that an object possesses by being in motion, or the ability for something to do work. The two equations that we learning in class involving kinetic energy (or KE) is KE = 1/2mv^2 and ∆KE = work. Another concept that we learned about is potential energy which is inversely proportional to kinetic energy. Potential is stored energy, for example a car at the top of a hill has potential energy. The equation used for potential energy is potential energy = mgh (mass x gravity x height).
Conservations of energy:
The equation for the conservation of energy is ∆KE = KE final - KE initial. This is similar to the conservation of momentum which is what we learned back in unit 4. The main question dealing with the conservation of kinetic energy is; Why do air bags keep you safe? And here is how you answer it.
KE = 1/2mv^2
∆KE = KE final - KE initial
You go from moving to not moving regardless of how you are stopped, there fore the ∆Ke is the same regardless of how you are stopped.
∆KE = work
Since the change in KE is the same regardless of what you hit, the work done on you is the same regardless of what you hit.
Work = force x distance (air bag)
Work = force x distance (no air bag)
Since the work is the same regardless of what is hit, the air bags increase the distance to stop you and thus reduces the force on you. Smaller force causes less injury.
Machines:
Like the conservation of energy which states that energy in a closed system is conserved and remains constant. Work when using machines is also conserved. For example, someone pushes a box up an inclined plane, and although by doing this you are using less force than if you just picked it up. But in both instances the work is the same. Pushing the box up the ramp is called the work in and lifting the box straight up is called the work out. And since work in = work out no matter how little force you put into pushing the box up the incline the work will always be the same because the distance has increased. The easiest way to understand this concept is by looking at an example problem: Emma pushes a 100N box up a ramp to put in the back of a U-haul truck, the ramp is 3 meters long and goes to a height of 1 meter. How much force does she use when pushing the box?
Work in = work out
force in x distance in = force out x distance out
(work in is using the ramp and work out is simply just picking up the box)
f x 4m = 100N x 1m
force = 25N
Machines help us decrease the force needed, but what you lose on the force you gain on the distance therefore work remains the same. When using some machines like a car if it is not 100% efficient the work out will be less than the work in, but this work didn't go anywhere, rather it is simply transformed into different forms of energy such as heat, sound, or smell. Although it is possible to have a smaller work out than a work it it is not possible to have more work out than work in because a machine can't be more than 100% efficient.
Problems that were difficult for me:
A car is moving at 20m/s speed and requires 10m to stop. How many meters will it take to stop if the speed of the car is 40m/s?
∆KE = KE final -KE initial
KE = 1/2mv^2
Since the car is moving twice the velocity at 40m/s we plug in 2 next to v
KE = 1/2 m(2v)^2
KE = 4 (1/2mv^2)
4 x KE
4 x work
work = f x d
4work = f x 4d
Since work is multiplied by 4 we know that the distance must be multiplied by 4 therefore to solve the stopping distance of the second car we must multiply 4 by the first cars stopping distance. The answer is 40meters. I overcame this problem by constantly looking over my notes on it and test. What eventually made the lightbulb click is that the only thing that really matters in the problem is how much faster the second car is going, the 20m/s and 40m/s shouldn't be included in the equation at all, the only take away from these speeds is that the second car was moving twice as fast.
This unit was harder for me than previous units, because all the formulas involved confuse me and causes me to mix up the major concepts and what equations go with which. By overcoming this confusion I made note cards to prevent me from mixing equations. My self confidence grows with each unit, and although learning a new topic can be overwhelming at times I usually understand it by the test. My goal for next unit is to come into conference more when I don't understand the homework problems, for this occurred a lot in this unit and I never did anything about it.
A connection that I made to real life was the potential energy on a roller coaster is conserved because the first hill is higher than all of the others.
Here is a podcast video I made with a group on the relationship between work and KE. Enjoy!
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