Dynamics solutions set 1

From the handout:

1. acceleration of block = .824 m/s2.
2. Tension in cable = 21000 N, lift force = 72500 N
3. Skydiver acceleration = 5.95 m/s^2 directed downwards
4. Slowing car: acceleration = 1.2 m/s^2, braking force = 1200 N
5. Woman must accelerate DOWN at 2.727 m/s^2
6. a) 980 N, b) 1010 N c) 965 N d) 980 N e) 0 N
7. a) 539 N b) 539 N c) 717 N d) 361 N e) 0 N


Book problems:
Question 11. Although the 2 kg rock has twice the gravity force, it also has twice the mass. By Newton's 2nd law, the acceleration must therefore be the same.

Problems:
12. T = 12960 N

13. accelerates DOWN at 3.5 m/s^2

15. If the thief is able to accelerate down the rope, the tension required to prevent him from falling will be less than his weight, and the "rope" will stay together.

Using the same logic as problem 5 from the handout, the acceleration must be downwards at 2.221 m/s^2 minimum for the rope not to break.

16. accelerates DOWN at .25 g, or at 2.45 m/s^2.

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.

30.
a) top cord tension = 58.8 N, bottom cord tension is 29.4 N
b) top cord tension = 68.4 N, bottom cord is 34.2 N

39.
a) Since the 40 N force is the minimum force required to get the box moving, the static friction force F_s = F_s,max = mu*F_n. Using a FBD and observing the box is in equilibrium until it starts to move, mu = F/mg = 0.8.

b) F - mu*mg = ma, so mu = (F - ma)/mg = 0.73

44. (After drawing a FBD of course!...) From Newton's 2nd, a = mu*g, and delta_x = -v_o^2/(2*mu*g) = 4.082 meters.