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103 PHYS - CH7 - Part2 Work due to friction If friction is involved in moving objects, work has to be done against the kinetic frictional force. This work is: W f = f k ⋅ d = f k d cos180o = − f k d 1 Physics Example r n r F θ f = µkn A block of mass m is pulled by a force F for a distance d on a rough surface as shown. What is the total work don on the block? r mg ŷ : Fnet , y = n + F sin θ − mg = 0 n = mg − F sin θ x̂ : Fnet , x = F cos θ − µ k n = F cos θ − µ k (mg − F sin θ) = F (cos θ + µ k sin θ) − µ k mg Wnet = Fnet , x ⋅ d N.B. find the work of each force and add the work, can you get the same answer? why WORK WORK IS IS SCALAR SCALAR QUANTITY QUANTITY Physics Dr. Abdallah M. Azzeer 2 1 103 PHYS - CH7 - Part2 Example 2.40 × 102 N force is pulling an 85.0-kg refrigerator across a horizontal surface. The force acts at an angle of 20.0˚ above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 8.00 m. Find (a)the work done by the pulling force, and (b) the work done by the kinetic frictional force. (a) W = F cos θ d = 1.8 × 103 J g (b) Wf = fkd cos θ fk = µk (mg - F sin θ) = 1.5 × 102 N F N 200 fk so Wf = -1.2 × 103 J d mg 3 Physics 7.5 Work & Kinetic Energy... A constant net Force r r r Work done by F ¨ W net = F ⋅ d = Fd Q F = ma , 1 ( 1 ⇒ Wnet = ma ⋅ v f 2 − vi 2 2a W net = 1 mv 2f − 1 mv i2 2 2 ) ( v f 2 − vi 2 v 2f = v i2 + 2ad ⇒ d = 2a K ≡ 1 mv 2 2 ) Wnet = K f − K i = ∆K {Net {Net Work Work done done on on object} object} == {change {change in in kinetic kinetic energy energy of of object} object} Physics Dr. Abdallah M. Azzeer 4 2 103 PHYS - CH7 - Part2 Kinetic Energy; Work-Energy Principle • • • • Energy ≡ The ability to do work Kinetic Energy ≡ The energy of motion “Kinetic” ≡ Greek word for motion An object in motion has the ability to do work • Net work an object = Change in KE. Wnet = ∆K The Work-Energy Principle Note: Wnet = work done by the net (total) force. Wnet is a scalar. Wnet can be positive or negative (because ∆KE can be both + & -) Units are Joules for both work & KE. Physics 5 Physics 6 Dr. Abdallah M. Azzeer 3 103 PHYS - CH7 - Part2 7.6 the nonisolated system COSERVATION OF ENERGY A book sliding to the right on a horizontal surface slows down in the presence of a force of kinetic friction acting to the left. The initial velocity of the book is vi, and its final velocity is vf. The normal force and the gravitational force are not included in the diagram 7 Physics Energy transfer mechanisms. (a) Energy is transferred to the block by work; (b) energy leaves the radio from the speaker by mechanical waves; (c) energy transfers up the handle of the spoon by heat; (d) energy enters the automobile gas tank by matter transfer; (e) energy enters the hair dryer by electrical transmission; and (f) energy leaves the lightbulb by electromagnetic radiation. Physics Dr. Abdallah M. Azzeer 8 4 103 PHYS - CH7 - Part2 We will be back to this point later Example 7.9,7.10 &7.11 9 Physics A Question • A box is pulled up a rough (µ > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces are doing work on the box? (a) 2 (b) 3 (c) 4 Solution Draw FBD of box: Consider direction of motion of the box Any force not perpendicular to the motion will do work: T v N N does no work ( ⊥ v ) T does positive work f does negative work mg does negative work 3 forces do work f mg 103 PHYS Dr. Abdallah M. Azzeer 5 103 PHYS - CH7 - Part2 READ READ EXAMPLES EXAMPLES 7.7 7.7 in in your your textbook textbook Example 7.8 A man loads a refrigerator onto a truck using a ramp. Ignore friction. He claims he would be doing less work if the length of the ramp would be longer. Is this true? Wnet = Wby man+ Wby gravity = 0 11 Physics Example A rescue helicopter lifts a 79-kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted from rest through a distance of 11 m. (a) What is the tension in the cable? (b) How much work is done by the tension in the cable (c) How much work is done by the person’s weight? (d) Use the work-energy theorem and find the final speed of the person. T g a) T-mg = ma , T= 8.3 × 102N d mg b) WT = Td = 9.13 × 103 J c) Ww = -mgd = -8.5 × 103 J d) WT + WW = 1/2 mvf2 - 1/2 mvo2 vf = 3.92 m/s Physics Dr. Abdallah M. Azzeer 12 6 103 PHYS - CH7 - Part2 How would you attack this problem? vy i = v0 sin θ ⎫ ⎪ a y = −g ⎪ 2 2 ⎬ v y f = v y i + 2a y ⋅ ∆y ∆y = −h ⎪ vy f = ? ⎪⎭ Eventually obtaining... v f = v02 + 2gh Independent of θ!... Given v0, θ, and h, what is the speed of the projectile the instant before it lands? OLD WAY 13 Physics WORK – ENERGY THEOREM ∑ W = Wg = K f − K i ∆y = −h Wg = −mg∆y = mgh Given v0, θ, and h, what is the speed of the projectile the instant before it lands? Physics Dr. Abdallah M. Azzeer 1 1 mgh = m v2f − m v02 2 2 2 2 v f = v0 + 2gh v f = v02 + 2gh NEW WAY 14 7 103 PHYS - CH7 - Part2 Example Two Two blocks blocks have have masses masses m m11 and and m m22 ,, where where m m11 >> m m22 .. They They are are sliding sliding on on aa frictionless when they they encounter encounter aa frictionless floor floor and and have have the the same same kinetic kinetic energy energy when long long rough rough stretch stretch (i.e. (i.e. m m >> 0) 0) which which slows slows them them down down to to aa stop. stop. Which Which one one will will go go farther farther before before stopping? stopping? (b) m2 (a) m1 (c) they will go the same distance m1 m2 103 PHYS Solution for the first mass m1 • The work-energy theorem says that for any object W = ∆K • In this example the only force that does work is friction (since both N and mg are perpendicular to the blocks motion). • The net work done to stop the box is – f d = -µ mg d. This work “removes” the kinetic energy that the box had: W= Kf - Ki = 0 - Ki m1 d 103 PHYS Dr. Abdallah M. Azzeer 8 103 PHYS - CH7 - Part2 for the first mass m2 • The net work done to stop a box is – f d = - µ mg d. ¾ This work “removes” the kinetic energy that the box had: ¾ W= Kf - Ki = 0 - Ki • This is the same for both boxes (same starting kinetic energy). µ m 2 g d 2 = µ m1 g d 1 Since m1 > m2 we can see that m2d2 = m1d1 d2 > d1 m1 m2 d1 d2 103 PHYS Example 0.075-kg 0.075-kg arrow arrow is is fired fired horizontally. horizontally. The The bowstring bowstring exerts exerts an an average average force force of of 65 65 NN on on the the arrow arrow over over aa distance distance of of 0.90 0.90 m. m. With With what what speed speed does does the the arrow arrow leave leave the the bow? bow? W net == F • d = 1 1 mv f2 − mv i2 2 2 vf = 39 m/s F d Physics Dr. Abdallah M. Azzeer 18 9 103 PHYS - CH7 - Part2 7.8 Power Power is the rate at which work is done by a force PAVG = W/∆t P = dW/dt Average Power Instantaneous Power The unit of power is a Joule/second (J/s) which we define as a Watt Watt (W) 1 W = 1 J/s = 1 kg.m2/s3 ( dW d r r = F⋅x dt dt r dxr r r P=F⋅ = F ⋅v dt P= ) In general, power is defined for any type of energy transfer. P = dE/dt Where dE/dt is the rate at which energy is crossing the boundary of the system by a given transfer mechanism. 19 Physics ∆W = F∆x cosθ F v θ ∆x = v∆t ∆W F∆x cosθ = Fvcosθ P= = ∆t ∆t Note: power × time = work, so work can be measured in units of kW.h (1 KWh = (103 Watt) × (3600 s) = 3.6 × 106 W s = 3.6 × 106 J =3.6 MJ.) • British units are hp (horse power) 1 hp = 550 ft ⋅ lb/s = 746 W Physics Dr. Abdallah M. Azzeer 20 10 103 PHYS - CH7 - Part2 Example 7.12 An elevator car has a mass of 1600 kg and is carrying passengers having a combined mass of 200 kg . A constant friction forces of 4000 N retards its motion upward, as shown in the figure. (a) What power deliver by the motor is required to lift the elevator car at a constant speed of 3 m/s? (b) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1 m/s2? (a) Mmax = 1800 kg, f = 4000 N v = 3 m/s (constant) ¨ a = 0 ∑F =T − f − Mg = 0 T = f + Mg = 4000 N + 1800 × 9.8 N = 2.16 × 104 N r r Power : P = T ⋅ v = Tv = 2.16 × 104 ( 3 ) = 64.8 kW ( ) (b) Left for you to try 21 Physics Example : Power Needs of Car Calculate Calculate the the power power neded neded far far the the car car (a) (a) to to climb climb aa hill. hill. (b) (b) to to pass pass another another car. car. ( a) ∑Fx = 0 F – FR – mg sinθ = 0 P = Fv ( b) Now, θ = 0 ∑Fx = ma F – FR= 0 v = v0 + at P = Fv Physics Dr. Abdallah M. Azzeer 22 11 103 PHYS - CH7 - Part2 Problem: determine the power after an elapsed time of 3.0 s P= mgh 50 kg × 9.8 m/s × 2.0 m = = 330 W ∆t 3.0 s Physics Dr. Abdallah M. Azzeer 2.0 m 50 kg g 23 12