The Physics Philes, lesson 14: The Ole Work-around
In which problems are solved, cars are pushed, and wood is hauled.
Last week I did my best to explain work, kinetic energy, and power. Now for the fun part. Drum roll please!
It’s time for the math!
Alrighty. Let’s start with a simple problem on work done by a constant force.
I bet you’re a pretty nice person, so when your friend calls upon you to help with their broken car, you show up with bells on. Your friend’s car is stalled! He needs you to push it out of the way. You exert a force with a magnitude of 210 N and you push it a distance of 18 m. Oh man! The car has a flat tire! So to get the car to go in a straight line you have to push it at a 30 degree angle. How much work did you do?
Do you remember what equation we use to find the amount of work done? That’s right! It’s the force exerted multiplied by the displacement. But in this question you had to push the car at a 30° angle. In this cases like this, we only need to know the force of the parallel component of the force. This means that the actual equation we need to use is W = FscosΦ. Luckily, we have all the information we need to solve this equation!
F = 210 N
s = 18 m
Φ = 30°
Now it’s just a straight forward case of plugging the values into the equation:
W = FscosΦ = (210 N)(18 m)cos 30° = 3.3 x 10^3 N·m = 3.3 x 10^3 J
That was pretty easy, right? Let’s do a similar problem, but this time we’ll calculate the scalar product. This time you’re pushing the car with a steady force of 160 N along the x-axis and -40 N along the y-axis. This displacement of the car is 14 m along the x-axis and 11 m along the y-axis. How much work did you do?
To find the scalar product, we need to use this equation:
y = 11 m
Again, it’s just a matter of plugging in the values:
W = (160 N)(14 m) + (-40 N)(11 m) = 2,240 J + (-440 J) = 1.8 x 10^3 J
Again, pretty easy stuff. You’ve probably got this stuff down pat. Now let’s use the work-energy theorem. The theorem can be used to calculate speed.
A tractor is hauling a sled loaded with wood 20 m. The total weight of the sled is 14,700 N. The total work done is 10,000 J. The initial speed of the sled is 2.0 m/s. How fast is the sled going after 20 m?
If you’ll remember, the work-energy theorem is total work = ΔK, or the final kinetic energy minus the initial kinetic energy. K = 1/2mv^2. The work done by the net force on a particle equals the change in the particle’s kinetic energy.
OK, what relevant information do we know about this situation? We know…
w = 14,700 N
initial speed = 2.0 m/s
total work = 10,000 J
final speed = ???
So our target variable is the final speed. We just need to find the initial kinetic energy and the final kinetic energy. Let’s work on the initial kinetic energy first. But whoops! We don’t have the mass of the sled. We only have the weight in newtons. But luckily, we can find the weight with the information we have.
Weight, if you’re remember, is the mass of an object times gravity.
w = mg
If we plug in the values, we get:
14,700 N = 9.8 m/s^2m
We need to rearrange the equation to solve for m:
m = 14,700 N/9.8 m/s^2 = 1,500 kg
Now we can solve for initial kinetic energy:
But that’s not our target variable. We need to find the final velocity.
Wee! You did it! You found the final speed! Good job! It’s not so bad. And now you know how to apply the work equation and the work-energy theorem. But don’t worry! I’ll be back next week to lay down some more energy knowledge. In the meantime, let me know of any errors and/or omissions you’ve seen in this post, or any post.
Featured image credit: makeshiftlove