Module 6—Work and Energy

Module 6: Lesson 1 Assignment

Remember to submit the answers to TR 1, TR 2, TR 3, and TR 4 to your teacher as part of your Module 6: Lesson 1 Assignment.

Try This

TR 1. A 2.20-N object is held 2.20 m above the floor for 10.0 s. How much work is done on the object?

TR 2. A 60.0-kg student runs at a constant velocity up a flight of stairs. If the vertical distance of the stairs is 3.2 m, what is the work done against gravity?

Hint: Use the formula Fg = mg.

TR 3. A 10.0-kg object is accelerated horizontally from rest to a velocity of 11.0 m/s in 5.00 s by a horizontal force. How much work is done on this object?

Hint: Use the following formula:

TR 4. A 90.0-N box is pulled 10.0 m along a level surface by a rope. If the rope makes an angle of 20.0° with the surface, and the force exerted through the rope is 75.0 N, how much work is done on the box?

Self-Check

SC 1. A 25.0-kg object is accelerated from rest through a distance of 6.0 m across a level floor in 4.0 s. If the force due to friction between the object and the floor is 3.8 N, what is the work done in moving the object? (Note: Fnet = ma, where a is the observed acceleration and Fnet = Fapp + Ff, where Ff is in the opposite direction of Fapp.)

When a bungee jumper is lifted at constant velocity via a crane platform, work is done against gravity, causing the jumper to have a given height relative to the ground below. In such a case, the lifting force applied to the jumper opposes his or her weight. Substituting weight (Fg) and vertical displacement (h) into the work equation produces a new equation for describing the work done in lifting an object against Earth’s gravitational field.

work = force × displacement
work = weight × vertical displacement
W = mgΔh
  
It is commonly accepted that if the object is lifted from the ground to a given height, the equation may be written as W = mgh. However, care should be taken to ensure that it is the change in height that is substituted into the equation.

Assuming that all of the work is done on the object being lifted vertically, the object gains an equal amount of gravitational potential energy. Therefore,

gain in potential energy = work done to lift it

Ep = mgΔh (or Ep = mgh, where h is the change in height)

1 J is equal to 1 N•m. Energy is a scalar quantity.

Gravitational potential energy is one form of potential energy involved in the bungee jumper-and-cord system. A second form of energy, called elastic potential energy, is also involved.

Read

Read about gravitational potential energy on pages 295 and 296 of your textbook.

Self-Check

SC 2.  Solve question 1 of “Practice Problems” on page 296 of your textbook.

Read

Read about the importance of the reference point used on pages 297 and 298 of your textbook.

Self-Check

SC 3. Solve question 3 of “Practice Problems” on page 298 of your textbook.