Assignment 2

Work and Energy

Read chapter 8 from Hewitt (pages 103 to 118) Good advice: use the review questions 1 - 20 on pages 119 to 120 as a "reading guide".

The very basic:

Question 1. How much gravitational potential energy is possessed by a 25 kg mass held at a height of 12 meters? The acceleration of gravity is -9.81 m/s/s by most references. You should generally round this value to -10 m/s/s.

E = mgh = 25•10•12 = 3000 J

Question 2. How much kinetic energy is possessed by a 12 kg mass traveling at 5 m/s?

E = 0.5•m•v•v = 0.5•12•5•5 = 150 J

Question 3. A spring is stretched by 2.5 meters from its "relaxed" length by a force of 5.0 N.

      3a. What is the spring constant (k) for this spring?

k = F÷∆x = 5 N ÷ 2.5 m = 2 N/m (That is, for every additional 2 N of force applied, the spring gets 1 meter longer).

      3b. How much energy is stored in the spring when it given an elongation (∆x) of 2.0 meters?

E = 0.5 •k•∆x•∆x = 0.5 •(2N/m)•2m•2m = 4 N•m or 4 J

Question 4. How much work is required to slide a refrigerator across a floor at constant speed if a 400 N force is required and the movement occurs over 4.0 m?

W = F•d = 400 N • 4 m = 1600 J

A little less basic:

Question 5. A pendulum is found to have a mass of 2.0 kg and its bottom edge is located 1.4 m above the ground at its highest point. How fast is it moving at its lowest point if it just barely misses touching the ground on its swing?

Conservation of energy requires that the potential energy at the highest point equal the kinetic energy at the lowest point as every Joule lost from potential on the way down must be transferred to kinetic. If it is virtually at the bottom at its lowest point it must then have 0 potential energy and a kinetic energy equal to the kinetic energy at the top. We can write: mgh (top) = .5 mv² (bottom). Solving for v we get v = sqrt(2gh) (since m cancels out). sqrt(2•10•1.4) = 5.3 m/s

Question 6. A ball is dropped from a height of 12.0 meters. How fast is it traveling when it passes the 6.0 meter mark on its way down?

Its gain in kinetic will equal its loss in potential: mg∆h = ∆mv•v and once again v = sqrt(2•g•h) = sqrt(2•10•6) = about 11 m/s.

Look at the physlet located at: http://www.physicsclub.net/physlets/physletprob/ch8_problems/ch8_3_work_energy/New_work_energy8_3_1.html

It may be easier to navigate from this sequence of links: physicsclub home page | Energy Basics | Ball in a Bowl

Question 7: How fast is the green ball moving at the bottom of the bowl?

Hint: Use the top of the green ball as a reference when you get the coordinates and remember that ∆Ep = ∆Ek

The ball starts at -2.6 m and falls to - 1.4 m a change: ∆y = 1.2 m. Its mass is 2 kg so ∆E = 2kg • 10 m/s/s • 1.2 m = 24 J

Kinetic = 24 J at bottom so V = sqrt(24J•.5÷2kg) = about 2.45 m/s

Look at the physlet located at: http://www.physicsclub.net/physlets/physletprob/ch8_problems/ch8_3_work_energy/work_energy_1.html (or navigate: home page, energy basics | work on a bowling ball)

Question 8: After trying all four animations, rank the four paths from most work by gravity to least work by gravity.

The net amount of work is independent of the path. Said another way, since the bowling ball started at the same low potential and finished at the same high potential each time, it necessarily follows that the work done by gravity was the same each time. For what it is worth -- the "hand" (the applied force) did positive work on the bowling ball equal to the product of the weight of the ball times the distance that the ball was moved. The amount of work was the same as the gain in potential energy from top to bottom.