Inertial Reference Frames and the Temple of Doom

Questions:
14.
The bag of groceries exerts a 40 N force(a) directed downwards (b) on the person (c).

16. While the forces are the same, the acceleration resulting from a "tug" may be different in both cases. The friction force, which provides an opposing force to the tension in the rope, largely determines whether either side will stay put.

22. By pressing horizontally, the normal force is increased. Since friction force is proportional to the normal force, the friction force directed up the wall will also increase.

Problems:
8. bullet acceleration = 21,875 m/s^2, force on bullet = 153 N.

17. Using g = 9.8 m/s^2, the maximum upwards acceleration is .557 m/s^2. Using g = 10 m/s^2, the maximum acceleration is .357 m/s^2. These should be solved by setting the tension to the maximum value, and then solving the equations of motion for acceleration.

33. You can approximate the two trains connected to a third that pulls it along by having three blocks in a row with two connecting strings:


The masses of the two left blocks are the same. Write Newton's 2nd for both blocks, and you should get that T1 = ma and that T2 - T1 = ma. This should help!

34. This is the same style as the pendulum problem from class. You should be able to obtain from your FBD (remember NOT to tilt your axes!) that T = mg/cos(theta) and that the acceleration is g*tan(theta).

The only catch is that you don't know acceleration. You DO know that it slows to a stop from 20 m/s in 5 seconds. Hint. Hint.

The answer is 21.8 degrees.

58.
a)
This problem works the same as having one block on the table with a string that holds a hanging block (like problem 81!).

For Cleo: T - Fk = ma and Fn1 - m1g = 0
For Figaro: m2g - T = m2a.

Solving algebraically, a = 0.098 m/s^2.
b)
Using kinematics, v0 = 0, a = 0.098 m/s^2, and delta_x = 0.9 meters.
Solving for delta_t = 4.111 seconds.

66. Force required = 21,000 N. Convert 90 km/h into m/s! displacement = 1.042 meters.

79.
Following the logic we used in class, we can derive that a = g tan(theta) = 4.663 m/s^2. From v = v0 + a*delta_t, with v0 = 0, and delta_t = 18 seconds, v = 83.9 m/s.