Work

Work relates to energy; whenever work is done, energy is transferred. It can change the amount of kinetic or potential energy an object has. Work can be done by applying a force for a certain distance. Based on these definitions, two equations can be written:

  - W = ΔE

  - W = F * Δx

where W is work, E is energy, F is force, and x is distance. Work is measured in newton-meters (N-m), also known as joules (J).

In class, we lifted and pulled different objects to demonstrate the concepts of work in the real world. The first item we lifted was a backpack.

Using a newton meter, we were able to figure out that the applied force had to be equivalent to 46 N in order to overcome the bag’s force of gravity. We lifted the bag over a distance of 1 meter. Because we know the force applied and the distance, we can use this to calculate the work done:

  - W = F * Δx = 46 N * 1 m = 46 N-m = 46 J

This answer also represents the change in energy because W = ΔE. In this case, both the kinetic and potential energy are increasing because it goes from rest to being in motion, and its height (relative to the ground) increases.

Next, we pulled my notebook across a table over a distance of 1 meter.

For this, the force of friction had to be overcome. Once again, we used a newton meter to determine that the applied force, which turned out to be 1.7 N. We can use this to calculate the work:

  - W = F * Δx = 1.7 N * 1 m = 1.7 N-m = 1.7 J

This time, only the kinetic energy is increasing because the notebook goes from rest to motion. Potential energy does not change because the height does not change throughout the motion. This also demonstrated that if you apply work, you will become much smarter; I transformed into a THREE-AP STUDENT!

Lastly, to demonstrate a situation in which no work was done, we applied force to an object that would not move.

This made the distance in “F * Δx” equal to zero. Because, anything multiplied by zero is equal to zero, the work done is zero (despite the amazing force):

  - W = F * Δx = 4109832785917230895710298375098127 N * 0 m = 0 N-m = 0 J

Another concept that is related to work is power. Power relates to work because it is defined as the rate at which work is done (or energy is dissipated). Using this definition, we can write the equation:

  - P = W/t

where P is power, W is work, and t is time. Power is measured in watts (W), which is equivalent to 1 joule (unit of work) over 1 second (unit of time). Another unit for describing power is horsepower (hp), which is approximately 746 watts.

In class, we tried to see if we could produce power that was greater than 1 hp. This was achieved by timing someone (Mr. 3-APs) running up a flight of stairs, determining the amount of work that was required, and then calculating the power to compare with 1 hp.

Data:

  - height of stairs: 2.62 meters

  - time to climb stairs: 2.34 seconds

  - mass of person: 63 kg

The force that must be overcome is the force of gravity:

  - Fg = m * g = 63 * 9.8 = 617.4 N

Work is the applied force (equivalent to the force of gravity) multiplied by the distance:

  - W = F * Δx = 617.4 * 2.62 = 1617.588 J

Power is the rate at which work is done (W/t):

  - P = W/t = 1617.588 / 2.34 = 691.277 W

Ultimately, our runner was not able to generate more than 1 hp because 691.277 W < 746 W = 1 hp. 

Through these various activities, we were able to see the real-life applications of the work and power equations. This activity was also important because it connected to some of our previously learned concepts such as forces.

Forces Practice Problems - 11

11. A student pulls a 35 kg box up a ramp that is inclined at 12 degrees. If the box starts from rest at the bottom of the ramp and is pulled at 185 N at an angle of 25 degrees with respect to the incline, what is the acceleration of the box along the ramp? Assume the coefficient of kinetic friction is 0.27.

List of all the given values:

• mass, m = 35 kg
• angle of incline = 12°
• applied force, Fapp = 185 N, at angle = 25°
• coefficient of kinetic friction, μk = 0.27

From this, a force diagram can be drawn:

We can use trigonometry to find Fapp*x and Fapp*y:

• cos 25 = Fapp*x ÷ 185 N, Fapp*x = 185 N * cos 25 ≈ 167.667 N
• sin 25 = Fapp*y ÷ 185 N, Fapp*y = 185 N * sin 25 ≈ 78.184 N

The force of gravity Fg can be found from the equation:

• Fg = mg = 35 kg * 9.8 m/s2 = 343 N

Using this Fg value, we can also find the “x” and “y” components of Fg using trig.:

• sin 12 = Fg*x ÷ 343 N, Fg*x = 343 N * sin 12 ≈ 71.314 N
• cos 12 = Fg*y ÷ 343 N, Fg*y = 343 N * cos 12 ≈ 335.505 N

Now that we know all the “y” forces, we can calculate the FN using the Fnet*y equation:

• Fnet*y = FN + Fapp*y - Fg*y = may
• because there is no acceleration, Fnet*y = 35 kg * 0 = 0
• so, we can set FN + Fapp*y - Fg*y = 0
• fill in known values: FN + 78.184 N - 335.505 N = 0
• FN = 335.505 N - 78.184 N = 257.321 N

Using the FN and the coefficient of kinetic friction (given), we can calculate the Ff:

• Ff = μk * FN = 0.27 * 257.321 N = 69.477 N

Finally, now that we know all the “x” forces, we can calculate the Fnet*x. This can then be used to find the acceleration:

• Fnet*x = Fapp*x - Ff - Fg*x = max
• Fnet*x = 167.667 N - 69.477 N - 71.314 N = 26.876 N
• we can set this equal to max: max = 26.876
• plug in mass: 35 kg * ax = 26.876, ax ≈ 0.768 m/s2

The final answer to this problem is 0.768 m/s2.

Free Fall

Free fall describes the motion of an object under the influence of only gravity.

Examples of free fall:

• a ball dropped in a complete vacuum - no air particles to provide air resistance
• a satellite orbiting Earth - nothing, but gravity, would affect the motion of the satellite. 
• a ball launched or thrown upwards in a vacuum - once the ball is launched, gravity is the only thing affecting its motion

Not examples of free fall:

• a ball dropped under normal conditions on Earth - the ball would experience air resistance
• a space shuttle (in space) which has its engines turned on - the engines would provide thrust, which would affect the motion of the space shuttle

Because of the negative acceleration of gravity, the position of an object dropped into free fall changes at increasingly larger rate. This is just like an object accelerating in linear motion. However, if the object is launched into the air, the position of the object will change at a smaller and smaller rate until it reaches the peak of its motion (zero velocity). After the peak, the position changes at an larger and larger rate.

In terms of velocity, the negative acceleration of gravity causes the velocity to steadily decrease whether the object is dropped or thrown.

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In class, we tried to demonstrate free fall as best as we could by dropping balls off of a staircase. It wasn't a "perfect" demonstration of free fall because we weren't dropping the balls in a vacuum, but we tried. :)

We recorded the dropping of a medicine ball using iPads and used an app to track the motion of the ball. We then created graphs based on the tracking information. Here are the results:

Video:

Tracking:

Graphs:

With best-fit line drawn in

These graphs demonstrate the motion described earlier. For instance, the slope of the position vs. time graph (top) becomes steeper as the ball drops. This shows that the ball is falling faster and faster, which means that the position is changing more and more.

The velocity vs. time graph shows the steadily decreasing velocity due to the negative acceleration. The slope of this graph also describes the acceleration of the ball which would be gravity in this case. Using the graphing app, we were able to create a best-fit line for the slope, and calculate the slope (acceleration of gravity). This yielded -9.708 (m/s^2), which is actually quite close to the known acceleration of gravity on Earth, -9.8 m/s^2. How close?

Percent error = |theoretical - actual yield|/(theoretical) * 100% = |9.8 - 9.708|/9.8 * 100% = 0.94% error

That's a very small percentage error despite the multiple potential sources of error. These include: 

• air resistance (goes against gravity --> smaller acceleration rate)
• video (doesn't show the whole drop, camera motion --> affect acceleration rate)
• accuracy of the tracking (we had to manually track the ball frame-by-frame --> affect acceleration rate)
• best-fit line (manually selected region --> affects how acceleration rate is calculated)

We also dropped a few other kinds of balls including a big tennis ball and a beach ball. Here are the acceleration rates of those:

• tennis ball: -9.795 m/s^2
• beach ball: -6.234 m/s^2

As you can see, the acceleration of the tennis ball is almost exactly the same as the acceleration of gravity. However, the beach ball's acceleration rate is off by quite a bit (~36% error). This is due to the higher air resistance experienced by the beach ball which creates a force opposite of the acceleration of gravity.

Overall, this demonstration was quite fun to experience, and I thought the tracking/graphing apps were pretty cool.