Work. Work = Force distance (the force must be parallel to movement) OR Work = (Force)(cos θ)(distance)

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1 Work Work = Force distance (the force must be parallel to movement) OR Work = (Force)(cos θ)(distance) When you are determining the force parallel to the movement you can do this manually and keep track of the direction of movement or you can use the general formula that W = Fcosθd. In this formula Fcosθ is equal to the force parallel to the movement. The drawback to using this generic formula is that you must pick the correct θ or the answer will be wrong. Alternatively, you can manually determine the vector that is parallel to the movement, decide if the movement is in the same direction as the force (positive force and therefore positive work) or in the opposite direction as the force (negative force and therefore negative work). Then use W = Fd with the appropriate sign for the force. Either method is acceptable, you will need to decide which works better for you. In addition to net work being equal to the net force times the distance moved, net work is equal to the change in kinetic energy of an object. The work done by gravity is equal to the change in the gravitational potential energy. Work done by a spring is equal to the change in the spring s potential energy. Each of these statements can be expressed as an equation: W net = KE f KE i (KE = ½ mv ) Kinetic Energy W gravity = PE f PE i (PE gravity = mgh) Potential Energy due to gravity W spring = PE f PE i (PE spring = ½ kx ) Potential Energy due to a spring Since energy is conserved, we can combine all of these into: Overall Work Equation (also called work energy theorem) ½ mv 1 + mgy 1 + ½ kx 1 F f d ± W ae = ½ mv + mgy + ½ kx KE i PE i spring i friction KE f PE f spring f (W ae = work from anything else) Many people find it easier to remember one equation than several. When using this equation, many of the terms are often zero and therefore the equation will simplify greatly. The units for work, kinetic energy and potential energy are Joules. 1 Joule = 1 Newton-meter. Depending on the problems often Nm are not converted to Joules until the end of the problem so that the canceling of units is more obvious. 1

2 Examples: i. Starting from rest a.5 kg block slides 3.0 m down a 50 incline. The coefficient of kinetic friction between the block and the incline surface is Determine the work done by a) the force of gravity b) the force of friction & c) the normal force New y F N F f F N F f 50 F gy Free body diagram F g New x F gx 50 F g W = Fd F g = Force of gravity m =.5 kg F N = normal force d = 3.0 m W g = work done by gravity = mgh (vertical height only) µ k = 0.33 W f = work done by friction = µ k F N d W n = work done by normal force = F N d cos 90 = 0 F g = mg = (.5 kg) (9.8 m/s ) = 5 N Σ F y = 0 F N F gy = 0 a. Work done by gravity F N = F gy = F g cos 50 W g = (F g sin 50) (d) = (.5 kg) (9.8 m/s ) (cos 50) = (5 N) (sin 50) (3.0 m) F N = 16 N W g = 57 Joules Force and movement are in same direction. Therefore, the answer is a positive number. b. Work done by friction *W f = -F f d OR W f = F f d (cos 180) = -µ k F N d = µ k F N d (cos 180) = -(0.33) (16N) (3.0) = (0.33) (16 N) (3.0 m) (-1) = -16 Joules = -16 Joules *Since force and movement are in opposite directions work is a negative number.

3 c. Work done by the normal force No component of the normal OR W n = F N d cos 90 = 0 force is parallel to the movement. Therefore W n = (0)(d) = 0 Since force and movement are perpendicular the normal force can t do work.. ii. A 50 kg man jumps off a 7 m cliff into the water. He continues to travel downward for another 3 m before coming to a stop. Using the work energy theorem, calculate the friction of water. 7 m All PE KE 3 m All KE lost to friction A. You can do this problem in two ways. A Convert PE to KE for the first 7 m of the man falling. Then convert KE to work of friction and solve for the force of friction. B Plug all the numbers to overall work equation and solve in one step. 1. First 7 m of falling m = 85 kg v 1 = 0 y 1 = 7 m v =? y = 3 m ½ mv 1 + mgy 1 = ½ mv + mgy ½ (85 kg) (0) + (85 kg) (9.8 m/s ) (7m) = ½ (85 kg) (v ) + (85 kg) (9.8 m/s ) (0) kg m 5831 = 4.5 kg (v ) s kg m 5831 s = v 4.5kg 137. m /s = v v = 11.7 m/s. Second 3 m where friction is stopping in motion m= 85 kg y 1 = 3 m v 1 = 11.7 m/s (this is v from part 1) y = 0 v = 0 ½ mv 1 + mgy 1 F f d = ½ mv + mgy 3

4 ½ (85 kg)(11.7 m/s ) + (85 kg) (9.8 m/s )(3.0 m) F f (3.0m) = Nm Nm F f (3.0 m) = = F f (3.0 m) 8317 Nm = F f 3.0 m F f = 77 N B. Use overall equation in step 1: M = 85 kg v 1 = 0 y 1 = 10 m v = 0 y = 0 d = 3 m (this is distance when friction is occurring) ½ mv 1 + mgy 1 F f d = ½ mv + mgy (1/) (85 kg) (0) + (85 kg) (9.8 m/s) (10 m) F f (3 m) = ½ (85 kg) (0) + (85 kg) (9.8 m/s) (0) 8330 F f (3.0m) = = F f (3.0 m) 8330 = Ff 3.0m F f = 777 N (close enough difference due to rounding off in method A) Problems: 1. A 100 kg box is moved 15 m by a 500 N force that is directed 37o upward. The µ k is What is the net work done on the box? Solution. A 0.50 kg object on a string is released from a height of 0.5 m. How fast is the object moving at the bottom of the swing? Solution 3. In a horizontal, frictionless system, a spring (k = 100 N/m) is compressed 0.00 m. What will be the speed of a kg object when the spring is released? Solution 4

5 Solutions: 1. A 100 kg box is moved 15 m by a 500 N force that is directed 37 o upward. The µ k is What is the net work done on the box? F f F N 500 N 37 0 F x = F applied(x) F g =mg F y m = 100 kg W net = F net d d = 15 m F net = F applied(x) - F friction F = 500 N F applied(x) = (F)(cos 37 o ) angle 37 o F friction = (F N )( µ k ) µ k = 0.00 F N = mg F y g = 9.8 m/s F N = (100 kg)(9.8 m/s ) - (500 N)(sin 37 o ) = 980 N N = N F net = (500 N)(cos 37 o ) (679.1 N)(0.00) = 400 N N = 64. N W net = (64. N)(15 m) = 3963 J = 4.0 x 10 3 J Return to Problems 5

6 . A 0.50 kg object on a string is released from a height of 0.5 m. How fast is the object moving at the bottom of the swing? m = 0.50 kg v 1 = 0 v =? y 1 = 0.5 m y = 0 g = 9.8 m/s 0.5 m ½ mv 1 + mgy 1 + ½ kx 1 F f d ± W ae = ½ mv + mgy + ½ kx 0 + mgy = ½ mv (0.50 kg)(9.8 m/s )(0.5 m) = ½ (0.50 kg)(v ) ()(9.8 m/s )(0.5 m) = v 4.9 m /s = v v =. m/s Return to Problems 6

7 3. In a horizontal, frictionless system, a spring (k = 100 N/m) is compressed 0.00 m. What will be the speed of a kg object when the spring is released? k =100 N/m x 1 = 0.00 m x = 0 m = kg v 1 = 0 v =? ½ mv 1 + mgy 1 + ½ kx 1 F f d ± W ae = ½ mv + mgy + ½ kx ½ kx = ½ mv (100 N/m)(0.00 m) = (0.350 kg)(v ) 4Nm = v 0.350kg 11.4 m /s = v v = 3.38 m/s 0.00 m Return to Problems 7

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