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Charles W.
Recent Harvard Graduate with Teaching Experience
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Physics (Thermodynamics)
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Question:

A ball of putty at temperature T = 300 K falls from a height of 1000 meters, deforming upon impact with the ground. The surrounding temperature is also T = 300 K. Assume no friction or air resistance, and that the ball makes no sound as it hits the floor. Important values:$$g = 9.81 m/s^2$$, specific heat: $$c_{putty} = 2.0*10^3 J/kg/K$$. Are there any energy transfer interactions across the boundary of the putty system during the fall? What is the temperature of the putty after it has just impacted with the floor before it has had a chance to transfer heat to the environment?

Charles W.
Answer:

As the ball of putty drops from 1 whole kilometer, there is obviously a transfer of energy occurring. Potential energy (in the form of height) is being converted into kinetic energy--making the ball of putty go faster as it drops. However, this is an internal change of energy. When thinking about transfer of energy across the boundary of the "putty system," we are asking the question of whether there is any outside influence that could alter the energy of the putty, such as a temperature differential or an outside force. As we have neither of those (notice the terminology "neglect air resistance" and that the room is of the same temperature), there is no energy transferred during the fall of the putty system. Now that we know there is no energy transfer, we know that the First Law of Thermodynamics is as follows:$(\Delta E_{net} = \Delta E_{potential} + \Delta U = Q - W = 0$)As we are interested in $$\Delta E_{potential}$$ and $$\Delta U$$, we simplify to$(-\Delta E_{potential} = \Delta U$)$$\Delta E_{potential}$$ is defined as $$mgh$$, due to the height of the drop, and $$\Delta U$$ is defined through our Energy Constitutive Relation, $$\Delta U = mc\Delta T$$. Now, when the putty hits the ground, there is no more potential energy, as the effective height of the putty is 0. Right before hitting the ground, we can say that all of the energy from being dropped from 1000 m has been transferred into kinetic energy:$(P.E. = K.E.$)After impact, all kinetic energy has been transferred to thermal energy, increasing the temperature of the putty system (remember, no energy is transferred to sound). This means that our potential energy has been effectively converted completely to thermal energy. We plug in what we know above into a final equation:$(-\Delta E_{potential} = \Delta U$)$(mgh = mc\Delta T$)$(9.81m/s^2*1000m = 2.0*10^3 J/kg/K * \Delta T$)$(\Delta T = 4.9 K$)$(T_{putty}=300 + 4.9 = 304.9K$)

Physics (Newtonian Mechanics)
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Question:

As you're getting up at the end of class, you accidentally hit the pencil on your desk, sending it flying horizontally off the edge. You've hit it with some gusto, and it doesn't even hit the ground until 2 meters from the edge of your desk. How fast was the pencil traveling as it left the edge of your desk? The top of your desk is exactly one meter off the ground, and you can neglect air resistance.

Charles W.
Answer:

Here, we should split the question into the two directions that the pencil is moving: away from the desk and down toward the ground. As there is no air resistance to worry about, the speed at which the pencil is moving does not create an external force on the pencil. This means that the horizontal motion of the pencil is independent of the vertical motion. This leads us to the important realization that the pencil will fall toward the ground at the same rate as a pencil that is simply dropped from rest at the same height (1 m). We calculate the time this takes by using our Newtonian equation of motion:$(x(t) = x_0 + v_0t + \frac{1}{2}at^2$)If we then rearrange this equation to see how long it takes for something to fall 1 meter from rest, our initial position is 0 m, our initial velocity is 0 m/s, and our acceleration is $$a=g = 9.81 m/s^2$$:$(1 m = \frac{1}{2}gt^2$)$(\frac{2m}{9.81m/s^2} = t^2$)$(\sqrt{\frac{2m}{9.81m/s^2}}=t$)We can then solve that $$t = 0.452s$$. Finally we move on to the horizontal motion of the pencil. As it is unaffected by the height at which we dropped the pencil, we know velocity has been constant, so the following equation applies:$(distance = velocity * time$)OR$(velocity = \frac{distance}{time}$) Plugging in what we know,$(velocity = \frac{2 m}{0.452s}$)$(velocity = 4.42 m/s$)

Geometry
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Question:

A budding entrepreneur, you are looking to create a new garden to plant both herbs and spices for the local farmers market. There is sufficient funding to rent 1500 feet of temporary chain-link fencing. The plan is to form two halves to your garden, with one shared fence running down the middle--the herbs will be planted on one side, and the spices on the other. What is the maximum area that the garden can obtain, and what are the dimensions of each of the two halves?

Charles W.
Answer:

To begin this problem, let's first get a better understanding of the layout of our fencing. To enclose the two sides of our garden with the most efficient use of our 1500 feet of chain link fence, we utilize a rectangle of fencing with a length of fencing the middle of the garden. A simplistic version of this looks like this:$(\fbox{ | }$)If we were to now label our fenced rectangle, the top and bottom would have length $$L$$ while the sides and middle portions all have length $$w$$. The total area of our garden is defined as $$Lw$$, and we can now create an equation for our perimeter:$(1500 ft = 2L + 3w$) (take note that this is different than our normal equation for perimeter, as there is the length of fence that splits the garden in two) If we now solve our perimeter equation for one variable (we choose $$L$$ in this case):$(1500-3w = 2L$)$(750 - 1.5w = L$)and we take this definition and plug it into our area equation:$(Area = (750-1.5w)w$)$(Area = 750w - 1.5w^2$)The form of this equation is known as a quadratic, and we can find the maximum area by simply finding the vertex of this equation. Remember, we organize quadratic equations with the highest power at the left, so our equation should be:$(Area=-1.5w^2 + 750w$) as we define $$a$$ as -1.5 and $$b$$ as 750. We now plug these in to our equation for the vertex:$(vertex = \frac{-b}{2a}$)This yields the final answer of:$(w = 250ft$)If we then plug this into our equation for perimeter, we get that our final length is:$(L = -1.5w +750$)$(L=-375 + 750$)$(L=375$)Finally, we compute area as being $$Lw$$, giving our final answer as 93,750 square feet.

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