6-3-98
Consider a person on a sled sliding down a 100 m long hill on a 30° incline. The mass is 20 kg, and the person has a velocity of 2 m/s down the hill when they're at the top. How fast is the person traveling at the bottom of the hill? All we have to worry about is the kinetic energy and the gravitational potential energy; when we add these up at the top and bottom they should be the same, because mechanical energy is being conserved.
At the top: PE = mgh = (20) (9.8) (100sin30°) = 9800 J
KE = 1/2 mv2 = 1/2 (20) (2)2 = 40 J
Total mechanical energy at the top = 9800 + 40 = 9840 J
At the bottom: PE = 0 KE = 1/2 mv2
Total mechanical energy at the bottom = 1/2 mv2
If we conserve mechanical energy, then the mechanical energy at the top must equal what we have at the bottom. This gives:
1/2 mv2 = 9840, so v = 31.3 m/s.
Now let's worry about friction in this problem. Let's say, because of friction, the velocity at the bottom of the hill is 10 m/s. How much work is done by friction, and what is the coefficient of friction?
The sled has less mechanical energy at the bottom of the slope than at the top because some energy is lost to friction (the energy is transformed into heat, in other words). Now, the energy at the top plus the work done by friction equals the energy at the bottom.
Energy at the top = 9840 J
Energy at the bottom = 1/2 mv2 = 1000 J
Therefore, 9840 + work done by friction = 1000, so friction has done -8840 J worth of work on the sled. The negative sign makes sense because the frictional force is directed opposite to the way the sled is moving.
How large is the frictional force? The work in this case is the negative of the force multiplied by the distance travelled down the slope, which is 100 m. The frictional force must be 88.4 N.
To calculate the coefficient of friction, a free-body diagram is required.
In the y-direction, there is no acceleration, so:
The coefficient of kinetic friction is the frictional force divided by the normal force, so it's equal to 88.4 / 169.7 = 0.52.
Being able to do work is not just what's important; how fast you can do work is also an important factor. Power is the measure of how fast work is done. Computers have more calculating power than we do; a sports car generally has a more powerful engine than an economy car. Power is the rate at which work is done and the rate at which energy is used. The unit for power is the watt (W).
An interesting calculation is the average power output of a human being. This can be determined from the amount of energy we consume in a day in the way of food. Most of us take in something like 2500 "calories" in a day, although what we call calories is really a kilocalorie; assuming we use up all this energy in a day (a reasonable assumption considering we'll have to eat tomorrow, too) we can use this as our energy output per day.
First, take the 2.5 x 106 cal and convert to Joules, using the conversion factor 4.18 J / cal. This gives roughly 1 x 107 J. Figuring out our average power output, we simply divide the energy by the number of seconds in a day, 86400, which gives a bit more than 100 W. In other words, on the average, we're just a little brighter than your average light bulb.
Power is work over time, and work is force multiplied by distance. Power can be written as:
Power : P = F s / t (F is the force in the direction of s, the displacement)
Displacement over time is velocity, so power can also be written in this form:
Power : P = F v (F is the force in the direction of the velocity)
Here's an example of when you might use this. Let's say you're riding your bicycle on a level road at a constant speed of 10 m/s. You're riding into a headwind, and you're burning up energy at the rate of 500 J/s. If you assume that 80% of this energy is going to overcome air resistance, how much force is the air exerting on you?
The power used to overcome air resistance is 80% of 500 W, which is 400 W. Assuming there aren't any other forces acting against you, then dividing this by your speed should give you the force the air exerts on you. This works out to 40 N.