This has the Potential to be a LOT of work! - Potential Energy

Hello everyone,

This week is going to require that we apply MOST of what we have learned thus far about Newton's Laws and Work/Energy. Working hard is required to really understand how it all comes together.

First, the solutions to the handout problems from class:

1.
a) Determine the magnitude and direction of the force that the spring exerts on the block. [45 N directed to the right.]
b) If the force holding the block against the spring is removed, calculate the work done by the spring on the block. [W_spring = 1/2kx² = 1/2(150 N/m)(0.3m)² = 6.75 J]
c) Calculate the speed of the block when it loses contact with the spring. [Using Work-Energy principle, W_net = KEf - KEi so v_f = 2.598 m/s.
d) If the compression fo the spring is doubled, how would your answer to (c) change? [final speed would double. ]

2.
a) Calculate the displacement of the spring from its equilibrium position. [.049 m]
b) Calculate the work done by gravity as the spring stretches.
c) Calculate the work done by the spring as the spring stretches. [.1801 J]
d) What is the net work done on the mass? [Wnet = 0 since the change in kinetic energy is zero!]
e) How much work was done in lowering the mass to the equilibrium position?

3.
a) Wnet = -24 J
b) Wspring = -1/2kx²
c) Wfric = -mu*mg*deltaX
d) -1/2kx² -mu*mg*deltaX = -24 J so deltaX = .325 m after solving numerically.

Homework Solutions/hints:
Questions:

5. Friction CAN cause an object to accelerate as in the textbook example of ripping out a tablecloth from under dishes. What does this mean about the change in KE? Wnet?

7. Using Uspring = 1/2kx²: (a) Spring 1 - displacement x = F/k. (b) Spring 2

8. KE = 1/2mv², where v is the speed and m is the mass for the object. Neither mass nor v² can be negative, so the answer is no.

9. The amount of work added the same in moving from B to C as compared with the work from A to B. The net work at point C, therefore, is twice the net work at point B. Using the Work-Energy principle, this means that the final speed is the square root of two times vb at point C.

Problems:
11. (a) 1.1*Mg (b) 1.1*Mg*h

14. Using area of a trapezoid: W = .5*(88N/m*.038m+88N/m*.058 m)*(.058m - .038m) = 0.08448 J

29. 0.337 meters

30. mgh = (6 kg)(10 m/s²)(1.2 m) = 72 J

33. a) mgh = (55 kg)(10 m/s²)(3100 m - 1600 m) = 825,000 J
b) minimum work required for the hiker to reach this new height IS the change in PE, or 825,000 J.
c) If the hiker also has a change in KE as compared with the KE at 1600 m, then the work could be more by the work-energy principle.