A man takes a ride on a ferris wheel. He boards a gondola at the very bottom of the ferris wheel (i.e. at ground level), which has a radius 10 meters. Can you determine how far he will have traveled along the ferris wheel's outer edge before he is 15 meters high in the air?
After the gondola that the man is riding in has moved through a total angle of 90 degrees, the gondola is 10 meters high in the air. Therefore, we need to figure out how much further (in degrees) he must travel before he moves 5 meters higher. If you draw a diagram of the situation, you can see that we want to find the angle of a right triangle with a hypotenuse of 10 meters and opposite side (height) of 5 meters. The angle is therefore arcsin (5 m / 10 m) = arcsin (1/2) = 30 degrees. Therefore, the man moves through a total of 90 + 30 = 120 degrees before he is 15 meters high in the air. To find the actual distance he has traveled, use the formula for arc length, i.e. s = r * theta. First convert 120 degrees into radians, i.e. (2pi/3). s = 10 m * (2pi/3) = 20.9 meters.
A homeowner is thinking of adding a plot of grass in his front yard, surrounded by fence. He has a total of P feet of fence. He wants to choose the dimensions of his plot of grass such that he will have the maximum amount of grass. Can you prove that he should choose a square-shaped plot in order to achieve his objective?
We want to find out what dimensions of a rectangle with a perimeter of P will have the maximum area. It turns out that the rectangle with maximum area for any given perimeter is always a square. We can show this as follows: Let L = length and W = width of the plot of grass. Let P = perimeter. For a rectangle, P = 2(L+W). Let the area = A. For a rectangle, A = LW. Let's solve for W in terms of L so that we can restate the function for area A in terms of only one variable. Let's use perimeter P to do this: P = 2(L+W) L+W = 0.5P W = 0.5P - L Therefore, the new equation for A in terms of L is: A = LW = L(0.5P - L) = (0.5P)L - L^2 Now let's find the extrema (max/min) of this function by taking the first derivative and setting it equal to zero: A(L) = (0.5P)L - L^2 dA/dL = 0.5P - 2L = 0 The solution to this equation is L = 0.25P. Before we proceed, though, we must confirm - is this a maximum or a minimum point? To find out, take the second derivative of A(L): A '' (L) = -2 Because the second derivative is negative, the function A is concave down for all values of L. Therefore, L = 0.25P must be a maximum. Now let's solve for W using the perimeter equation and L=0.25P obtained above: P = 2L + 2W P = 0.5P + 2W W = 0.25P In other words, the length and width are exactly the same for a rectangle with maximum area of perimeter P, i.e. L = W = 0.25P.
A 2 kg cubical box is resting on a flat, horizontal table and its left side is pressed against a spring with a spring constant of k = 100 N/m. The spring is compressed to 70 cm from its equilibrium point. The table is covered with sandpaper that has a kinetic coefficient of friction of 0.6. When the spring is released, how far will be box slide across the table before coming to rest?
You can solve this problem using work and conservation of energy. At the start, before the spring is released, the box has a total potential energy due to the compressed spring of E = 0.5 * k * x^2 = 0.5 * (100 N/m) * (0.7 m)^2 = 24.5 J. As the box travels across the sandpaper, the friction force between it and the box acts in a direction opposite to its motion, thereby doing negative work on the box until it comes to rest. The force of friction does a total amount of work of -24.5 J on the box before it comes to rest (i.e. so that its final energy is zero). Since work = force * displacement and the force of friction is F = -mu * N = 0.6 * mg = -0.6 * (2 kg) * (9.8 m/s^2) = -11.76 N, the total displacement of the box is (-24.5 J)/(-11.76 N) = 2.08 meters.