Work and Power
Work is a transfer of energy and is responsible for power. The equation for work is, Work = Force x distance. Work is measured in Joules (J). For work to be done on an object, the force and distance must be parallel to one another.
In this image, the weight lifter is lifting a weight with a total mass of 50 kg. He is lifting the weight a distance of 2 meters. How do we solve for the work being done here? We already know the distance, so first, we must solve for the force.
Force = mass x gravity
= 50 x 10
= 500 N
Now that we know the force being used to lift the weight, we can find the work being done.
Work = Force x distance
= 500 x 2
= 1000 J
I mentioned earlier that work is responsible for power, but what is power? Power is how quickly work is done, or Power = work/time. Power is measured in Joule/seconds, other known as Watts (W). A Watt is 1 Joule over 1 second. So let's calculate the amount of power generated by the weight lifter. Suppose it took him 5 seconds to lift the weight.
Power = work/time
= 1000/5
= 200 W
Fun Fact! You've probably heard the term horsepower in reference to cars and other modes of transportation. 1 Horsepower actually equals 746 Watts!
Work and Kinetic Energy
Kinetic Energy is the energy of movement. The equation for Kinetic Energy is, KE = 1/2mv^2. How does KE relate to work? Work is the change in KE. By solving for the change in KE, you also solve for work. The equation for change in KE is, ∆KE = KEfinal - KEinitial
Let's solve some practice problems to better understand work and KE relationship.
1. A 20kg car accelerated from 20m/s to 30m/s in 5 seconds. In that time, it traveled 100m.
a. What was the change in energy the car experienced.
This problem is asking us to find the change in kinetic energy. To do this, we must first find the initial and final kinetic energies.
KEinitial = 1/2mv^2 KEfinal = 1/2mv^2
= 1/2 (20)(20^2) = 1/2 (20)(30^2)
= 1/2 (20)(400) = 1/2 (20)(900)
= 1/2 (8000) = 1/2 (18000)
= 4000 J = 9000 J
Once we solve for the initial and final KE, we can solve for ∆KE
∆KE = KEfinal - KEinitial
= 9000 - 4000
= 5000 J
Another example of this concept can be seen in my group's video.
b. How much work was done?
Since work = ∆KE, we know that the work done is 5000 J.
2. A car is moving at come speed and requires 10m to stop. How many meters will it take to stop if the speed of the car is tripled? Why?
The first KE = 1/2mv^2
The second KE = 1/2m(3v^2)
= 9 (1/2mv^2)
Since the car's velocity is tripled and this number is squared, it's KE is 9 xs greater than its initial KE.
Because the car's KE is 9 xs greater, it's work is also 9 xs greater. Work = force x distance, so the distance must be 9 xs greater.
d = 9(10)
= 90m
Conservation of Energy
Along with kinetic energy, there is something called potential energy. Potential energy is the energy of position. As long as an object is at some height, moving or not moving, it has potential energy. The equation for potential energy is, PE = mgh. Anytime an object moved, energy is conserved, meaning it doesn't change, even if something changes its movement. Energy is conserved, because anytime PE changes (increases or decreases), KE changes as well (increases or decreases in response).
Take this swinging pendulum for instance. At the beginning of its swing, the ball has the creates PE, because it is at the greatest height. As the ball lowers and PE decreases, KE increases as the ball speeds up in response to the decreasing PE. KE is greatest at the bottom of the ball's swing. On the right side of the pendulum, the PE will be the same as it was in the beginning, because it is at the same height, and PE is the energy of position.
This conservation of energy suggests that the energy before will equal the energy after. Yet, we often hear about energy being lost. In these situations, energy isn't being lost, but is being produced in unused forms. Some of these forms include light, sound, and heat. One example of this is a car. Much of the energy from gasoline is produced as a vibration of the car, a rumbling of the engine, and the engine heating up. The amount of energy actually being used in a system is known as its efficiency. Percent efficiency is found by solving for work out/ work in.
Machines
Machines help us use our energy more efficiently by reducing the amount of force needed to move an object. In this unit, we addressed simple machines. A prime example of a simple machine is the inclined plane. As previously mentioned, work = F x d. The inclined plane, and all other simple machines, increase the distance an object moves, in turn decreasing the force needed to move it. Although the force exerted is decreases, the work will remain the same as it would have been lifting an object over a short distance.
In this image, a man is pushing a 200 N object up an inclined plane of 12m. The ramp has a vertical height of 8m. The work that would be done if the man lifted the 200 N straight up 8m is called the workout. The work done pushing the weight up the 12m ramp is called the workin. We can use the following equation to solve for the amount of force needed to push the object up the ramp.
workin = work out
Fin x din = Fout x dout
F x 12 = 200 x 8
12F = 1600
F = 133.3 N
As you can see, the workout = the work in, but the force required to push the object up the ramp is smaller than the force required to lift it straight up.
My problem solving skills, effort, and learning…
In this unit, I have completed all of my assignments on time and prepared well for all assessments. My group filmed a video about work and kinetic energy that was informative and went over what we covered in class on the subject. I finally learned how to embed videos in my blog, as can be seen above. In this unit, I struggled with the conservation of energy, but coming into conference period clarified my confusion.
My goals for the next unit…
1) Continue turning in work on time
2) Continue preparing well for tests and quizzes
3) Make my group video more interesting
