Chapter 5: Work Work is the transfer of energy that occurs when a force makes an object move. Energy is always transferred when work is done. Two conditions for doing work 1. Force must make the object move. 2. Movement must be in the same direction as the force Example: When you carry books you are exerting an upward force to lift the books, but since the motion is horizontal there is no work being done.

W = Fd W F d W = work, units of Joules F = force, units of Newtons d = distance, units of meters W F d Examples: 1. A force of 100 N is used to move a refrigerator a distance of 5 meters. How much work was done? W = Fd W = (100 N)(5 m) W = 500 J 2. 200 J of work is done to move a box a distance of 10 meters. What force was used? F = W  d = W/d F = (200 J)/(10 m) F = 20 N

P = W/t Power: the amount of work that can be done in one second P = power, units are Watts (W) W = work, units are Joules (J) t = time, units are seconds (s) P = W/t Example: You and a friend are having a race to see who can push a box of books up a ramp the fastest. The boxes weigh the same (the force is the same), and the distance is the same but the winner has done the work in less time. The winner has more power.

P = W/t W P t Examples: 1. You do 900 J of work pushing a sofa for 5 s. What was your power? 2. A 5 W motor is used for 10 s. How much work was done? 3. A 150 W motor does 900 J of work. How long was it running? P =W / t P = (900 J)/(5 s) P = 180 W W = Pt W = (5 W)(10 s) W = 50 J t = W/P t = (900 J)/(150 W) t = 6 s

P = E/t When energy is transferred, power can be calculated using: P = power, units are Watts (W) E = energy, units are Joules (J) t = time, units are seconds (s) P = E/t Example: A 100 W light bulb is left turned on for 6 hours. How much energy was used? Convert 6 hours to seconds: (6 hours)(60 min/hr) = 360 minutes (360 min)(60 s/min) = 21600 seconds E = Pt E = (100 W)(21600 s) E = 2.16 x 106 J

By changing the applied force. Machine: a device that makes doing work easier... By changing the applied force. By changing the distance over which the force is applied. By changing the direction of the applied force.

For an ideal machine: Win = Wout Input force: the force that is APPLIED TO the machine, Fin Ex: Pushing the jack handle up and down. Output force: the force APPLIED BY the machine, Fout Ex: The force from the jack lifts the car. Input work: the work that is done by Fin, symbolized by Win Output work: the work that is done by Fout, symbolized by Wout For an ideal machine: Win = Wout

MA = Fout Fin Mechanical Advantage: Machines make doing work easier. The amount that they make it easier to do work is the mechanical advantage (MA). MA = Fout Fin Fout = output force, units are N Fin = input force, units are N Ideal Mechanical Advantage (IMA): Is the MA for a machine without friction. IMA = Fout Fin IMA = din dout

By applying a force of 50 N, a pulley system can lift a box with a mass of 20 kg. What is the mechanical advantage of the pulley system? MA = Fout/Fin Fin = the force applied to the machine Fin = 50 N Fout = the force supplied by the machine to lift the mass Fout = mg = (20 kg)(9.8 m/s2) Fout = 196 N MA = Fout/Fin = (196 N)/(50 N) MA = 3.9

The EFFICIENCY of a machine can be found by dividing the output work by the input work. x 100 Win Wout Lubricants increase efficiency by reducing the amount of friction between two surfaces. Less friction means that less of the Win is converted into heat energy.

What is the efficiency of a machine if you do work on the machine at a rate of 1200 W and the machine does work at a rate of 300 W? Efficiency = (Wout/Win)*100 The problem gives power values not work values. Assume that the time involved is 1 second. Win = Pt = (1200 W)(1 s) = 1200 J Wout = Pt = (300 W)(1 s) = 300 J Efficiency = (Wout/Win)*100 = (300 J/1200 J)*100 = 25%

First class lever Fout Fin Fulcrum Output arm Input arm length of input arm length of output arm IMA = = Lin Lout

IMA = Lin Lout IMA = 5 cm / 5 cm = 1 IMA = 2 cm / 8 cm = 0.25 IMA =

2. Make a line 2 cm from one end. Label the line OUTPUT. 1. Take a strip of posterboard and cut it in half. Tape the long edges of the two pieces together. 2. Make a line 2 cm from one end. Label the line OUTPUT. OUTPUT

3. Slide your lever toward the edge of the desk until it just BEGINS to tip. Mark this line. Mark this point OUTPUT LINE Desk 4. Write INPUT at line in middle. 5. Weigh paper strip and record weight.

6. Center dime on OUTPUT line 6. Center dime on OUTPUT line. Slide strip toward edge of table until it BEGINS to tip. Mark the point at the edge of the table FULCRUM 1. 7. Repeat with nickel and quarter. Mark their tipping points FULCRUM 2 (nickel) and FULCRUM 3 (quarter). 8. Measure from the OUTPUT line to FULCRUM 1. Record this as OUT1. 9. Measure from the INPUT line to FULCRUM 1. Record this as IN1. 10. Repeat for FULCRUM 2 AND FULCRUM 3.

You should end up with a data sheet that looks like: Measured using balance Mass of strip: _____________ g Mass of dime: _____________ g Mass of nickel: ____________ g Mass of quarter: ___________ g IMA = Lin/Lout Calc coin wt = IMA * mass of strip