The Power of the People

Hope you all enjoyed your election day off. I'm looking forward to some very interesting results tonight as polls close. This has the makings of quite a historic day for all sorts of reasons.

We will start tomorrow by going over the Pre-Lab questions, and then get straight to making measurements with the pull-back cars. Please make sure you go through the steps and think about what we will be doing in class so everything goes smoothly.

We rushed through the last steps to the Mars Rover problem, so here are those solutions, along with the book problem solutions:

Question 7.
e) Since the batteries hold 5.4 x 10⁵ J, and the power required to drive is 10 W, we can figure out how long the batteries will hold up under these conditions:

P = W/delta_t so delta_t = 54,000 s. Since the rover drives at 6.7 x 10^-3 m/s, we can calculate the distance traveled (as we are given the rover land speed) using delta_x = v*delta_t = 361.8 m

f) .010% of the power goes against drag, so this means that .001 W goes against drag. Since the speed is constant, we can use P = F*v to calculate the friction force F = .149 N.

58. P = W/delta_t so delta_t = (285 kg)(9.8 m/s^2)(16.0 m)/1750 W = 25.54 seconds

64. Using Wnet = delta_KE, Wnet = 715.4 J. P = W/delta_t = 357.7 W

67. By drawing a FBD, we can obtain that the force required to move at constant speed must be F = mg*sin(6). The displacement moving up the hill is delta_x, and cos(theta) = 1, since the force is directed along the surface of the hill.

Our equation for power becomes P = F*delta_x/delta_t = mg*sin(6) * v, so v = P / (mg*sin(6)).

The only hiccup is that we need to convert hp (horsepower) to Watts. We can do this using the information on page 169 that says that 1 W = 746 W. This means that P = 186.5 W.

Substituting, v = 2.6 m/s.

71. Since the terminal velocity is 30 m/s, we can assume this is the paratrooper's speed before hitting the snow.
a) Wnet = delta_KE = -36,000 J
b) Wnet = F_average*delta_x*(-1) = -36000 J. delta_x = 1.1 m, so F_average = 32,700 N
c) W_ncf = delta_K + delta_U = .5*(80 kg)*(30 m/s)^2 - 0 + 0 - (80 kg)*(9.8 m/s^2)*(370 m) = -2.54 x 10⁵ J