Chapter 10 Work and Energy

Objectives: Describe work in terms of force and displacement, using the definition of the scalar product. Solve problems involving concept of work. Distinguish between the resultant work and the work of a single force. Define the spring constant and calculate the work done by a varying spring force.

Three things are necessary for the performance of work: q x There must be an applied force F. There must be a displacement x. The force must have a component along the displacement.

Work = Force component X displacement Definition of Work Work is anything that can exert a force through a distance. a scalar quantity Work = Force component X displacement   Unit: 1 Nm = 1 Joule (J)

Work can be: Positive, when force applies in the same direction as the displacement. F x Negative, if the force is opposite displacement direction

If a force does not affect displacement, it does no work. Work done by any force that is at right angles with displacement will be zero (0). The force F exerted on the pot by the man does work. F W The earth exerts a force W on pot, but does no work even though there is displacement. If a force does not affect displacement, it does no work.

Resultant Work or Net Work Resultant work is the algebraic sum of the individual works of each force. F x f Example: F = 40 N, f = -10 N and x = 4 m Work = 120 J Work = (40 N)(4 m) + (-10 N)(4 m) OR Example: Work = (F - f) x Work = (40 - 10 N)(4 m)

Note: Work is positive since Fx and x are in the same direction. Example 1: A lawn mower is pushed a horizontal distance of 20 m by a force of 200 N directed at an angle of 300 with the ground. What is the work of this force? 300 x = 20 m F = 200 N F Work = (F cos q ) x Note: Work is positive since Fx and x are in the same direction. Work = (200 N)(20 m) Cos 300 Work = 3460 J

Graph of Force vs. Displacement Assume that a constant force F acts through a parallel displacement Dx. Force, F The area under the curve is equal to the work done. F x1 x2 Area Work = F(x2 - x1) Displacement, x

Example for Constant Force What work is done by a constant force of 40 N moving a block from x = 1 m to x = 4 m? 40 N Force, F Displacement, x 1 m 4 m Area Work = F(x2 - x1) Work = (40 N)(4 m - 1 m) Work = 120 J

Work of a Varying Force Our definition of work applies only for a constant force or an average force. What if the force varies with displacement as with stretching a spring or rubber band? F x F x

Hooke’s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. F = -kx F x m The spring constant k is a property of the spring given by: K = DF Dx

Work Done in Stretching a Spring Work done ON the spring is positive; work BY the spring is negative. F x m From Hooke’s law: F = kx Work = Area of Triangle x F Area = ½ (base)(height) = ½ (x)(Favg ) = ½ x(kx) Work = ½ kx2

Compressing or Stretching a Spring Initially at Rest: x m Compressing Stretching Two forces are always present: the outside force Fout ON spring and the reaction force Fs BY the spring. Compression: Fout does positive work and Fs does negative work (see figure). Stretching: Fout does positive work and Fs does negative work (see figure).

General Case for Springs: If the initial displacement is not zero, the work done is given by: x1 x2 F x1 m x2

Power of 1 W is work done at rate of 1 J/s Power is defined as the rate at which work is done: (P = dW/dt ) Power of 1 W is work done at rate of 1 J/s

Units of Power One watt (W) is work done at the rate of one joule per second. 1 W = 1 J/s and 1 kW = 1000 W One ft lb/s is an older (USCS) unit of power. One horsepower is work done at the rate of 550 ft lb/s. ( 1 hp = 550 ft lb/s = 746 W)

Recognize that work is equal to the change in kinetic energy: Example 4: A 100-kg cheetah moves from rest to 30 m/s in 4 s. What is the power? Recognize that work is equal to the change in kinetic energy: m = 100 kg Power Consumed: P = 1.22 kW

CONCLUSION: Chapter 10 - Work