2. UNIT 4: WORK, ENERGY & POWER PART I

3. WHAT IS WORK? A force causing displacement Time is not a factor----can be fast or slow Force must be applied in the same direction as the displacement of the object or at some angle theta (Θ) Work = force x displacement x cos Θ If force and displacement are in the same direction then Θ is 0 (W = Fd) Measured in N m or Joules

4. POTENTIAL ENERGY Energy that is stored and waiting to be used later Energy of an object due to its condition or position

5. POTENTIAL ENERGY Potential energy due compression or expansion of an elastic object.

6. POTENTIAL ENERGY Potential energy stored within the chemical bonds of an object

7. POTENTIAL ENERGY

8. KINETIC ENERGY Energy resulting from motion of an object. Work must be done to cause an object to be in motion. Work = change in kinetic energy W = ΔKE KE = 1/2 mv 2 (measured in Joules)

9. EXAMPLE A 105-g hockey puck is sliding across the ice. A player exerts a constant force of 4.50-N over a distance of 0.150 m. How much work does the player do? What is the change in the puck’s energy? What is the change in velocity of the puck? W = (4.50 N)(0.150 m) = 0.675 J ΔKE = W = 0.675 J 0.675 J = (0.5)(0.105 kg)v 2 = 3.6 m/s

10. EXAMPLE When a golf club strikes a 46-g golf ball, the ball picks up 43 J of KE. A constant force of 2300 N is applied. Over what distance is the club in contact with the ball? W = Fd  d = W/F d = 43 J/ 2300 N = 0.019 m

11. EXAMPLE A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes a 25° angle with the horizontal. How much work does the sailor do on the boat if exerts a force of 255 N on the rope? W = Fdcos Θ W = (255 N)(30.0 m)(cos 25°) = 6933.3 J

12. EXAMPLE Will N. Andable and Ben Pumpiniron are in the weight room. Will lifts a 100-lb. barbell over his head 10 times in 1 minute while Ben lifts a 100-lb. barbell over his head 10 times in 10 seconds? Which student does the most work? Answer:

13. EXAMPLE Calculate the work done by a 2.0-N force (directed at a 30° angle to the vertical) to move a 500 gram box a horizontal distance of 400 cm across a rough floor at a constant speed of 0.5 m/s.

14. TOTAL MECHANICAL ENERGY

15. POWER Power is the rate at which work is done. Measures how quickly energy is being transferred Power = work/time or P = W/t Measured in Watts (W) which equals J/s 1 horsepower (hp) = 750 W Substituting Fd for W  P = Fd/t Substituting v for d/t  P = Fv

16. EXAMPLE A net force of 2800 N accelerates a 1250- kg vehicle for 8.0 s. The vehicle travels 80.0 m during this time. What power output does this represent? P = W/t = Fd/t P = (2800 N)(80.0 m)/8.0 s = 28,000 W Can also be expressed in kiloWatts (28 kW)

17. EXAMPLE Remember Will N. Andable and Ben Pumpiniron? Will lifted a 100-lb. barbell over his head 10 times in 1 minute while Ben lifted a 100-lb. barbell over his head 10 times in 10 seconds? Which student delivers the most power?

18. EXAMPLE If Nellie Newton is doing chin-ups during her physical fitness test and lifts her 42- kg body a distance of 0.25 m in 2 seconds, what power is delivered by Nellie’s biceps? F g = mg = (42 kg)(9.8 m/s 2 ) = 411.6 N P = Fd/t = (411.6 N)(0.25 m)/2 s = 51.5 W

19. EXAMPLE A squirrel (mass of 1 kg) does push-ups by applying a force to elevate its body by 5 cm. Determine the number of push-ups a squirrel must complete in order to do 1 Joule of work. If he does them in 4 s, determine his power output. W = Fd = mgd = (1 kg)(9.8 m/s 2 )(0.05 m) = 0.49 J The squirrel must do ____ push-ups. P = W/t = 1.0 J/ 4 s = 0.25 W

20. SIMPLE MACHINES Eases the load by changing either the magnitude or direction of a force to match the force or capability of the operator. Mechanical advantage (MA) is a ratio of the resistance force (F r ) exerted by the machine to the effort force (F e ) exerted by the operator. MA = F r /F e

21. TYPES OF SIMPLE MACHINES Pulley Inclined Plane Lever Wheel & axle Screw Wedge

22. SIMPLE MACHINES Ideal Mechanical Advantage (IMA) is the ratio of the displacement of the effort force (d e ) to the displacement of the load (d r ) IMA = d e /d r In theory, all machines would be 100% efficient….but in reality they are not Efficiency (e) = MA/IMA x 100 or work output divided by the work input [W o /W i x 100]