Am I done? Great. I've got other things to do.
We went over the, ahem...
Unbalanced force Particle Model:
Newton's Second Law Review Problems
We went over the, ahem...
Unbalanced force Particle Model:
Newton's Second Law Review Problems
...worksheet.
Using a new, groundbreaking technique where the students put the answers on the blackboard, we answered problems 3, 4, and 5 in class:
3.
Using a new, groundbreaking technique where the students put the answers on the blackboard, we answered problems 3, 4, and 5 in class:
3.
We will use this equation to find unbalanced Forces:
I hope you enjoy my childish scrawl.
M meaning our mass (which in this case is 0.15 kg) and A meaning our acceleration.
But, we need to find our A, which we measure in m/s/s.
However, you'll notice that we have an issue with measurements. We'll use measurements in m and seconds (s), but the question starts out with a measurement in mi/hr.
Homey don't play that.
None of you will really get this reference.
Which brings me to why I left Chemistry class: Dimensional Analysis.
This horrid mathematical device was the general tool in converting the terrible mi/hr to our darling meters per second (m/s).
Just in case you've forgotten how to do this, I'll try to explain this in a way that won't make me have another panic attack.
So, we set up 100 mi/hr in the upper left. According to our sheet, 1 mi is about 1600 m. So, we set the ratio (1600 m/1 mi) up next to it. Mi goes on the bottom because it's sort of like multiplying fractions, and the old measurements are supposed to cancel. We need to get rid of that measurement to convert our given.
So, we can successfully convert our mi to m by multiplying 100 and 1600. But, now we need s.
Because hr is below on the left, it has be be above on the right, again so it can cancel out. below it goes 60 minutes (min). The ratio is (1 hr/60 min). But, we want s, so we add another ratio (1 min/60 s), so we can convert our measurement to seconds.
I might have made this more confusing.
Multiply 100 by 1600. Then divide it by 60 twice. You about 44.4 m/s.
We'll use that to find the acceleration with this equation:
Velocity (V) squared equals Initial Velocity (Vo) squared plus 2 times A times change in position (ΔX).
Our V is 44.4 m/s. Our Vo is assumed to be zero. Our ΔX is 1.5 m minus zero m, which is 1.5 m.
So, with our acceleration, now we go back to our first equation and plug in our numbers:
98.6 Newtons (N) are exerted on the ball to throw it at 100 mi/hr.
4.
After staring from rest, a 0.50 kg ball falls 30 m. It has contact with the ground for 2 milliseconds (.002 s) before bouncing back up and reaching 20 m. We need to find the force exerted on the ball as it hit the ground.
Because we have a mass again, we'll use our Funbal = MA equation.
We need to find the acceleration as it hit the ground, which would be between our two velocities.
Oh my God, NO!!
Yes, two velocities. There is when it is falling initially and after it bounces on the ground. Luckily, we can use the same equation from before:
Or, at least I hope it is.
So, I put 9.8 in for A.
"But then, we have acceleration! Why do we need to figure out what it is if we have it?"
Well, very bold student, we have acceleration, while the ball is in the air, while the acceleration due to the force of gravity acts upon it, not when it hits the ground.
So, we get 24.25 m/s from the first one and 19.8 m/s from the second. We take these two and stick them into another equation:
Good ol' change in Velocity (ΔX) over change in Time (ΔT).
If we consider the bounce upwards to be "moving forward," then the ball dropping would be considered "moving backwards." That's why the velocity is negative now.
Doing this equation leads us to the ungodly acceleration of 22,025 m/s/s.
Plugging our numbers back into our Funbal equation:
According to maths, our answer is 11,012.5 N.
5.
Another sled problem. That's all I should have to say.
There's a pair of kids in a sled. One of them is having too much fun. With the sled, their mass is 100 kg. A man with no muscle mass is pulling the sled by way of a rope at a 25° angle. The tension on the rope is 120 N. The friction force is 15 N. We have to find the acceleration.
We can use the same formula we have been using, but this time we have to change it to solve for A instead:
We have a mass (100 kg), so we just need our unbalanced Force. We make a Force diagram:
X is our force forward, which we need to find for the unbalanced force.
The reason the normal force doesn't equal the force of gravity is because of that 50 N over on the right. The rope is pulling the sled to the right and up, to the ground isn't pushing up against the sled as much as it would is it were at rest.
To find X, we use Trigonometry:
We subtract 15 N from 108.76 N and get 93.76 N, our unbalanced force.
We plug it into our adapted Funbal equation:
The acceleration is .94 m/s/s, the child was having way too much fun, and this entry is finished.
Using this information, we had to find the acceleration. First, we made a force diagram for both the weight and the cart. The force diagram for the cart looked like this: 
and the force diagram for the weight looked like this:
To find the acceleration we used the formula acceleration=Funbalanced/mass. The unbalanced force is 1 N, and the mass is 0.6 kg, so the equation would look like: acceleration=1 N/0.6 kg. When you solve the equation, the acceleration is 1.67 m/s/s.
