Discover the answer to this question in this chapter.

Transcription

1 Erwan, whoe ma i 65 kg, goe Bungee jumping. He ha been in free-fall for 0 m when the bungee rope begin to tretch. hat will the maximum tretching of the rope be if the rope act like a pring with a 100 N/m contant? Dicover the anwer to thi quetion in thi chapter.

2 Origin of ork The tudy of different imple ytem ince antiquity howed that the weight multiplied by the vertical diplacement (P y) had a particular ignificance. Beginning with Archimede, who noted that P y i the ame for two mae on a lever in equilibrium if there i a diplacement, up to Galileo, who howed that P y determine the velocity of a pendulum, thi quantity wa ued to explain many phenomena. Decarte ummed up, in 1637, the ue of thi quantity. The invention of thee device i baed on one principle, which i that the ame force that can lift a weight, for example, 100 pound, at the height of two feet, can alo lift a weight of 00 pound at the height of one foot, or a weight of 400 pound at the height of a half-foot, and o on Becaue it i the ame to lift 400 pound at the height of one-half foot, and to lift 00 at the height of one foot, and the ame to lift 100 at the height of two feet. (arning: Decarte ue the term force even if he wa not talking about force. At that time, there wa no conenu on what force wa and the term wa ued for almot everything in mechanic.) From thi principle, the operation of everal machine ued at the time wa explained, uch a lever, pulley, inclined plane, pendulum and many other. ith the arrival of Newton mechanic, thi principle went omewhat in the background but did not diappear. It wa realized later that a very intereting reult i obtained in Newton mechanic if the force i multiplied by the diplacement. Several name were given to thi quantity, uch a mechanical power, quantity of action or dynamic effect until Gapar Corioli gave it it current name (work) in 186. The ork Done by a Contant Force If a contant force act on an object moving along a traight line, then ome work i done. Thi work i the calar product of the force and the diplacement. The ork Done by a Contant Force Acting on an Object Moving in a Straight Line () = F coθ or = F 019 Verion 8 ork

3 i the diplacement of the object and θ i the angle between the diplacement and the force. du-phyic.org/phyic180/phyic195/topic/chapter7.html The unit ued to meaure work i the Nm. Another name wa given to thi unit: the joule (J). The Unit of ork: Joule 1 J = 1 Nm = 1 kgm ² ² If there are everal force acting on an object, the um of the work done by each of thee force i the net work. The Net ork Done on an Object ( net ) = = net 1 3 Example hat i the net work done on thi box if it move 3 metre toward the right? Let calculate the work done by each of the force acting on the box. The work done by the 0 N force i 1 = 0N 3m co 90 = 0J Verion 8 ork 3

4 The angle i 90 becaue the force i directed downward and the diplacement i toward the right. The work done by the 10 N force i = 10N 3m co180 = 30J The angle i 180 becaue the force i to the left and the diplacement i toward the right. The work done by the 50 N force i The angle i 60 a hown in the figure. The net work i therefore Here ome commentary. 3 = 50N 3m co 60 = 75J net = = 0J + 30J + 75J = 45J 1) F and are never negative. Thee are alway the magnitude of the force and the diplacement. ) There i no need to reolve the force into component when calculating the work with F coθ. The magnitude of the force and the diplacement mut be put in the formula, not the component. 3) The work can be poitive, negative, or zero. A F and are alway poitive, the value of the angle determine the ign of the work. A the coine i poitive for an angle maller than 90 and negative for an angle between 90 and 180, the ign of the work i a follow. 4) The angle i alway poitive. Alway ue the mallet angle between the force and the diplacement, and there no ign for thi angle. The angle will therefore alway be between 0 and Verion 8 ork 4

5 Example 8.1. A 5 kg object lide 10 m downhill on a 30 lope. The coefficient of friction between the lope and the object i 0.. hat i the net work done on the object between thee two intant? 03o-and-i-conne-q To find the work done by the force, the force acting on the object mut be found firt. The force are: 1) The weight. ) A normal force. 3) A friction force. The work done by the weight i N = mg coθ = 5kg m co 60 = 45J g kg The angle between the force and the diplacement i 60 according to the figure. The work done by the normal force i = F 10m co 90 = 0J N N The angle between the force and the diplacement i 90 according to the figure. The work done by the friction force i = µ F 10m co180 f c N The angle i 180 becaue the friction force i directed uphill and the diplacement i directed downhill. To find thi work, the magnitude of the normal force mut be found. ithout going too much into detail (becaue many example of calculation of the normal force on a lope were done in the previou chapter), the equation of the y-component of the force i 019 Verion 8 ork 5

6 The normal force i then ( ) F = F + mg in 60 = 0 y N N F = mg in 60 = 5kg 9.8 in 60 = 4.435N N and the work done by the friction force i, therefore, kg = µ F 10m co180 = N 10m 1 = 84.87J f c N Thu, the net work i = + + = 45J + 0J J = 160.1J net g N f There i another way to calculate the work ince, according to what you have learned in vector algebra, the calar product can be calculated from the component of the vector F and. The ork Done by a Contant Force on an Object Moving in a Straight Line () = F = F x + F y + F z x y z Example F = i + j + k N act on an object moving from the point (1 m, m, 3 m) to The force ( ) the point (5 m, 6 m, -3 m). hat i the work done on the object? Going from (1 m, m, 3 m) to (5 m, 6 m,- 3 m), the component of the diplacement are Therefore, the work i x = 4m y = 4m z = 6m = F x + F y + F z x y z = 3N 4m + 4N 4m + 5N 6m = 1J + 16J 30J = J 019 Verion 8 ork 6

7 The ork Done if F or θ Are not Contant If the magnitude of the force or the angle between the force and the diplacement change, the diplacement mut be divided into part where the force and the angle are contant. The work done i obtained by umming the work done in each of thee part. The ork Done if F or θ change () Example = F coθ contant F and θ or = F contant F and θ A force directed toward the right act on an object whoe diplacement i 6 m toward the right. The force i 5 N over a ditance of 5 m and then 3 N over a ditance of 1 m. hat i the work done on the object between thee two intant? A the force change, the diplacement mut be divided into two part. The work done during the firt part i 1 = F coθ = 5N 5m co 0 = 5J The work done during the econd part i Therefore, the total work i = F coθ = 3N 1m co 0 = 3J = + 1 = 5J + 3J = 8J 019 Verion 8 ork 7

8 Example A 1 N force acting toward the right act on an object that move following the path hown in the figure. hat i the work done on the object? A the angle between the force and the diplacement change, the diplacement mut be divided into part. The work done during the firt part (motion toward the right) i 1 = F coθ = 1N 10m co 0 = 10J The work done during the econd part (upward motion) i Therefore, the total work i = F coθ = 1N 10m co 90 = 0J = + 1 = 10J + 0J = 10J But what hould be done if the force or the angle are continuouly changing? It eem that the diplacement could not be divided into egment where the force i contant then. Actually, it can be done. The diplacement i divided into very hort ditance, o mall that they become infiniteimally mall. The work done on uch a ditance i d = F dx If the work done during thee mall diplacement are then added, the total work i obtained. ork Done on an Object () (Mot General Formula) = F coθ d or = F d 019 Verion 8 ork 8

9 There an integral in thi formula becaue thi i what you get when you um infiniteimally mall quantitie. Properly peaking, thi i a line integral a it i the reult of a calculation along a line (ince the object pae from one place to another following a path, which i a line). It i poible to calculate thi kind of integral for three-dimenional trajectorie, but thi i beyond college curriculum. Only diplacement along an axi (which will be denoted x) will be conidered here. In uch a cae, the infiniteimal work i F d = F dx x and the work i ork Done by a Variable Force on an Object Moving Along the x-axi (the Object Goe from x to x ) = x x F dx x Example A variable force F = (3 N/m x + N) act on an object going from x = 1 m to x = 3 m. hat i the work done on the object between thee two intant? The work i 3m 1m N ( 3 ) = x + N dx m N 3 m x = + N x ( m) 3m 1m ( m) N 3 3 N 3 1 m m = + N ( 3m) + N ( 1m ) = 16J 019 Verion 8 ork 9

10 Graphical Repreentation of ork Let re-examine the cae of the work done by a contant force with a graph of force veru poition. A horizontal line then repreent the force. If an object move from x to x, the work i F x. But F x i alo the area under the curve of F between x and x ince thi i a rectangle with height F and bae x. Thi concluion remain valid even if the force i variable. hen Fdx wa calculated, the area of a thin rectangle whoe height i equal to the force at thi place and whoe width i equal to dx wa calculated. hen thee infiniteimal work are ubequently added, the um of all the area of thee rectangle between x and x wa obtained. Thi eem not to give exactly the area under the curve becaue there are mall miing piece at the top of the rectangle, but actually, the rectangle are much thinner (infiniteimal) and much more numerou (infinite) than what can be een in the figure. The mall miing piece then diappear completely, and the area under the curve i exactly equal to the work done on the object. The work done on an object i the area under the curve of the force acting on the object a a function of poition. A the work can ometime be negative, thi area can alo be negative. A in calculu, the area i poitive if it i above the x-axi and negative if it i below the x-axi. However, thee ign are inverted if the object move from a higher value of x to a maller value of x. Thi come from the fact that the ign of an integral change if the limit of an integral are inverted. 019 Verion 8 ork 10

11 en.wikipedia.org/wiki/integral The ork Done by a Spring ith the definition of work obtained previouly, the work done by a pring on an object can be calculated. A the force made by a pring i F = -kx, the work done by a pring on an object moving from x to x i p x ' x' kx kx kx = ( kx) dx = = x x Thi can be implified to obtain ork Done by a Spring k p = x x ( ) Thi calculation could alo have been done by calculating the area under the curve of the force. Since the graph of -kx i a traight line with lope - k paing through the origin, the area under the curve between x and x correpond to the grey area in the following figure. 019 Verion 8 ork 11

12 Thi area i Area = (Area of the triangle ranging from the origin to x ) (Area of the triangle ranging from the origin to x) x' F ' x F Area = Since the magnitude of the force i kx, the area become x kx x kx Area = kx kx = A thi area i under the x-axi, the work i negative. Therefore, the work i p kx kx = k = x ( x ) Thi i the ame reult a the one obtained with the integral. Example A box liding on a horizontal urface hit a pring, which compree the pring to 0 cm. The pring contant i 1000 N/m. hat i the work done by the pring on the box between thee two intant? The work done by the pring i k p = x x ( ) At intant 1, the pring i not compreed. Therefore, x = 0 m. At intant, the pring i compreed 0 cm. Therefore, x = 0 cm. 019 Verion 8 ork 1

13 Thu, the work done i (( 0. ) ( 0 ) ) N 1000 m p = m m = 0J It i normal for the work to be negative in thi cae, becaue the force exerted by the pring on the box i directed toward the left while the diplacement of the box i toward the right. The ork Done by a Spring and the ork Done on a Spring Be careful, there a difference between the work done by the pring and the work done on the pring. The work done by the pring force i the work done on an object puhed or pulled by the pring. It correpond to the work that ha already been calculated. Force made by a pring on an object Fp = kx ork done by the pring on an object p = ( x x ) The work done on the pring i the work that an object or omeone mut do to compre or tretch the pring. However, according to Newton third law, the force that an object mut exert on a pring to tretch or to compre it ha the ame magnitude a the force that the pring exert on the object, but in the oppoite direction. Thi change of direction remove the minu ign in the formula of the force, and thu alo change the ign of the work. Therefore, the force and the work are faraday.phyic.utoronto.ca/pvb/harrion/flah/tutorial/flahphyic.html Force made by an object on a pring F = kx k = x x ork done by an object on a pring ( ) k Proof of the ork-energy Theorem Let now examine why it can be ueful to calculate the work done on an object. 019 Verion 8 ork 13

14 The proof tart with the definition of net work. = F d net where and are the poition at two different moment. Since F net = ma according to Newton econd law and the acceleration i the derivative of the velocity, thi equation become net net = ma d dv = m d dt v d = m dv v dt A the derivative of the poition i the velocity, thi become net v = mv dv v But ince the net work can be written a If thi integral i done, the work i d v = d v v = v dv + dv v = v dv ( ) ( ) v mv dv v m 1 net = = v v d v 1 v net = m ( ) d v v 1 v = m v v 1 = m( v v ) 1 1 = mv mv ( ) The term found in the lat line were given the name kinetic energy. 019 Verion 8 ork 14

15 Kinetic Energy (E k ) Ek = 1 mv The work i then 1 1 net = mv mv = E E k k Finally, the work-energy theorem i obtained. ork-energy Theorem net = E k Therefore, the net work done on an object give the change in kinetic energy. Thi theorem allow u to eaily find the change of peed of an object. The ign of net work thu ha a new meaning. 1- If the net work i poitive, then the kinetic energy increae, and the peed of the object increae. - If the net work i negative, then the kinetic energy decreae, and the peed of the object decreae. 3- If the net work i zero, then the kinetic energy i contant, and the peed of the object i contant. Example 8..1 A 5 kg box lide 10 m down on a 30 lope. The coefficient of friction between the lope and the box i 0.. hat i the peed of the box after the 10 m lide if the box wa initially at ret? 03o-and-i-conne-q Verion 8 ork 15

16 The work-energy theorem will be applied here for the motion between thee two intant: Intant 1: The box i at the top of the lope. Intant : The box located i 10 m down the lope. net Calculation The net work done on the box between thee two intant wa already calculated in an example of the previou ection (example 8.1.). The net work wa net = 160.1J E k Calculation ith an initial velocity of zero, the change in kinetic energy i 1 1 Ek = mv mv 1 1 = =.5kg v Application of the ork-energy Theorem m ( ) 5kg v 5kg 0 net = E 160.1J =.5kg v v = k m Example 8.. Erwan, whoe ma i 65 kg, goe Bungee jumping. He ha been in free-fall for 0 m when the bungee rope begin to tretch. hat will the maximum tretching of the rope be if the rope act like a pring with a 100 N/m contant? The work-energy theorem will be applied here for the motion between the two intant hown in the figure. In thi figure, d i the tretching of the rope Verion 8 ork 16

17 E k Calculation A Erwan ha no peed at both intant, the change in kinetic energy i net Calculation 1 1 Ek = mv mv 1 1 = 65kg = 0J m m ( ) kg ( ) To calculate the net work done between the two intant, the force acting on the object mut be found. The force are: 1- A 637 N weight directed downward. - The force made by the rope (like a pring) directed upward. Thi force will act only after an initial downward diplacement of 0 m. The work done by the weight i The work done by the pring i = F coθ g g ( ) ( ) = 637N 0m + d co 0 = 637N 0m + d k p = x x ( ) At the beginning, the tretching i zero (x = 0), wherea it i d at the end (x = d). Therefore, the work done by the pring i p k = N = 50 d ( d 0 ) m Thu, the net work i = + net g p N ( ) = 637N 0m + d 50 d m 019 Verion 8 ork 17

18 Application of the ork-energy Theorem net = E N ( ) 637N 0m + d 50 d = 0J N 50 m d 637N d 1, 740J = 0 The olution of thi quadratic equation i d = 3.56 m. (The other olution d = 10.8 m correpond to a compreion of the rope, which i impoible here.) The rope thu tretche 3.56 m, for a total fall ditance of m. k m Note that the olution of thi problem would have been much longer if it had been made uing only Newton econd law becaue it would have been neceary to do an integral to find the change of peed while the rope tretche. A the force i not contant, the acceleration i not contant and all the contant acceleration motion formula cannot be ued here. Thi i why an integral would have been required to find the diplacement. The following example i a road afety example. In it, it i hown that the braking ditance increae with the quare of the peed. So don t go too fat, becaue if you do, you ll need a lot of ditance to top. Example 8..3 hat i the minimum braking ditance of a car going at 90 km/h if the coefficient of tatic friction between the tire and the road i 0.9? The work-energy theorem will be applied here for the motion between thee two intant: Intant 1: the car i moving at 90 km/h. Intant : the car i topped, farther down the road. fr.depoitphoto.com/577683/tock-illutration-car.html 019 Verion 8 ork 18

19 E k Calculation A the peed change from 90 km/h to 0 km/h, the kinetic energy variation i net Calculation 1 Ek = mv 1 = mv To calculate the net work between thee two intant, the force acting on the car mut be found. The force are: 1) The weight, directed downward. ) A normal force directed upward. 3) A breaking force (a friction force) oppoed to the diplacement. The work done by the weight and the normal force are both zero becaue thee force are perpendicular to the velocity and, therefore, to the diplacement. Only the friction force doe work on the car here. Thu 1 mv = = F coθ net f f A the force i in the oppoite direction to the diplacement, thi become Application of the ork-energy Theorem = F co180 = F net f f net = E 1 Ff = mv 1 Ff = mv k A the tatic friction force mut be maller than µ F N F µ F f N The following equation mut alo be true. F µ F f N 019 Verion 8 ork 19

20 Since 1 Ff = mv, the equation i 1 mv µ FN A the normal force i equal to the weight in thi cae, the equation become 1 mv µ mg 1 mv µ mg v µ g Therefore, the minimum breaking ditance i v min = µ g ( 5 ) = = 35.43m Thu, the minimum braking ditance of a car i proportional to the quare of the peed. m N kg = min v µ g Thi increae of the breaking ditance with the peed can be een in thee clip. The formula alo indicate that the coefficient of friction alo influence the braking ditance. A larger coefficient of friction mean a maller the topping ditance. The coefficient of friction between the road and formula 1 tire being much greater than for an ordinary car tire, the braking ditance of a Formula 1 i much maller. It i alo important not to lock the wheel while braking. To minimize the breaking ditance, you mut brake a hard a poible, without locking the wheel. If the wheel are locked, then they lide on the road. Thi mean that you then have kinetic friction intead of tatic friction. A the coefficient of kinetic friction i maller than the coefficient of tatic friction, the braking ditance increae if the wheel are locked. Thi i why anti-lock 019 Verion 8 ork 0

21 braking ytem (ABS) are ued: they prevent the locking of the wheel. The other major utility (in fact, the main) of thi ytem i to enable the driver to maintain control of hi vehicle while breaking. Indeed, it i impoible to teer a car when the wheel are locked. Even if you turn the wheel, the car doe not turn then. By preventing the wheel from locking, the ytem allow the driver to keep control of hi car even during an intene braking. Definition The average power i the work done divided by the time required to do thi work. Average Power ( ) P = t The unit of power i J/. The name watt () wa given to thi unit. Unit of Power: att J 1 = 1 = 1 kgm ² ³ Another unit i ometime ued for power: the horepower (hp), which i worth 746. Actually, a hore can provide much more power than one horepower. Thi 746 i the reult of an etimate made in the 19 th century of the average power of a hore when it work without too much effort. Thi mut not be confued with the cheval-vapeur, which i worth only 736! Unit of Power: Horepower 1hp = 746 The intantaneou power i calculated exactly a average power, but by conidering the work done in a very hort time (an infiniteimal time). The intantaneou work i thu Power (P) d P = dt 019 Verion 8 ork 1

22 Power correpond, therefore, to the rate at which work i done. Uing what i known about work, the power can be written a F coθ P = = (If the force i contant) t t d Fd coθ P = = dt dt Thi become Average and Intantaneou Power P = Fv co θ = F v (If the force i contant) P = Fv coθ = F v Graphical Interpretation The formula d P = dt obviouly mean that, on a graph of work done on an object a a function of time, the lope i the power. ork from Power If the power i contant, the following equation P = t 019 Verion 8 ork

23 mean that the work can be obtained from the power with ork from Power if the Power I Contant = P t If the power i not contant, the following formula mut be ued d P = dt to obtain the work done for infiniteimal time. d = Pdt Then all thee work are added to get ork from Power if the Power I not Contant = Pdt Graphically, thi mean that the work done i equal to the area under the curve of the power a a function of time. 019 Verion 8 ork 3

24 Example Example A 1000 kg (including the paenger) elevator move upward with a teady peed of 3 m/. If the frictional force oppoing the motion of the elevator i 4000 N, what i power (in hp) of the engine which pulled the elevator? The force made by the engine i found with the um of the vertical force. F = mg + F F = 0 y engine f engine F = 9800N N engine F = mg + F F engine = 13,800N f Therefore, the power i P = F v coθ engine engine m = 13,800N 3 co 0 = 41, 400 = 55.5hp The angle i 0 ince the force and the velocity are both directed upward. Example 8.3. The power of the engine of a 1000 kg car i 1 hp when the car move on a horizontal urface with a contant peed of 80 km/h. hat i the power of the engine if the ame car move uphill on a 10 lope with the ame contant peed of 80 km/h? Let tart by looking at what happen on the horizontal urface. Thi will allow u to know the frictional force acting on the car when it i travelling at 80 km/h. The force on the car are hown in the figure. 019 Verion 8 ork 4

25 fr.depoitphoto.com/577683/tock-illutration-car.html (The normal force i actually acting on all four wheel. The point of application of the force of the engine i at the contact of the wheel and the ground ince it i the friction force between the wheel and the road which make the car accelerate.) If the peed i contant, then F = F F = 0 x engine f F f = F The force of the engine can be known ince the power of the engine i known. The force i P = F v coθ engine engine engine m 1hp 746 = F. co0 F engine engine = 40.8N The angle i 0 a the force and the velocity are both directed toward the right. e now know that the magnitude of the friction force acting on the car i 40.8 N when it i moving at 80 km/h. If the car now move uphill on a 10 lope, the force hown in the figure act on the car. fr.depoitphoto.com/577683/tock-illutration-car.html 019 Verion 8 ork 5

26 To find the power of the engine, the force made by the engine mut be found. It i found with the um of the x-component of the force. F = F F + mg co100 = 0 x engine f F = F mg co100 engine f Since the car i moving at the ame peed a it wa on the horizontal urface, the frictional force i the ame (40.8 N). The equation now become Therefore, the power i F = F mg co100 engine f N = 40.8N 1000kg 9.8 co100 = 40.8N N = 104.6N P = F v coθ engine engine m = 104.6N. co 0 = 46, 768 = 6.7hp kg It can be een that the required engine power increae much when the car move up the lope. Some car could not even achieve thi feat due to lack of power. The famou Chevette of the 80, having only a 53 hp power (bae model), would not have been able to move up thi lope at 80 km/h. Example A 4 kg object i ubjected to a force whoe power i given by en.wikipedia.org/wiki/chevrolet_chevette 6 ² 4 P = t + t a) Knowing that the object wa at ret at t = 0, determine the peed of the object at t = 3. The work done on the object i 019 Verion 8 ork 6

27 = ( 6 t 4 ² ) 0 Pdt = + ² 3 3 ² t t 0 = + ( ) ( ) 3 = = 7J t dt Thu, the peed i = E 1 1 7J = mv mv 1 v = 6 7J = 4kg v 0 m k b) Knowing that thi i a motion along the x-axi, determine the force acting on the object at t = 3. The force will be found with P =Fvcoθ. For a motion in one dimenion, the angle can only be 0 or 180. The power at t = 3 i P = t + t 6 ² 4 ² ( ) = = 66 Thu, the formula for the intantaneou power give P = Fv coθ m 66 = F 6 coθ Since the power i poitive, the angle mut be 0. Therefore, the magnitude of the force i m 66 = F 6 co 0 F = 11N 019 Verion 8 ork 7

28 The ork Done by a Contant Force on an Object Moving in a Straight Line () = F coθ or = F or = F x + F y + F z x y z The Net ork Done on an Object ( net ) = = net 1 3 The ork Done by a Contant Force on an Object Moving in a Straight Line () = F coθ Contant F and θ or = F Contant F and θ The ork Done on an Object () (Mot General Formula) = F coθd or = F d ork Done by a Variable Force on an Object Moving along the x-axi (the Object Goe from x to x ) = x x F dx x 019 Verion 8 ork 8

29 The work done on an object i the area under the curve of the force acting on the object a a function of poition. ork Done by a Spring k p = x x ( ) Kinetic Energy (E k ) Ek = 1 mv ork-energy Theorem net = E k Average Power ( ) P = t Power (P) d P = dt Average and Intantaneou Power P = Fv co θ = F v (If the force i contant) P = Fv coθ = F v 019 Verion 8 ork 9

30 The power i the lope on a graph of the work a a function of time. ork from Power if the Power I Contant = P t ork from Power if the Power I not Contant = Pdt The work i the area under the curve of the power on a graph of the power a a function of time 8.1 Definition of ork 1. hat i the work done by Mutafa in thi ituation? cnx.org/content/m4147/latet/?collection=col11406/latet 019 Verion 8 ork 30

31 . Honoré puhe a 30 kg crate up an 8 lope over a ditance of 5 m. The crate ha an acceleration of 1 m/² toward the top of the lope, and there i a frictional force of f = 70 N oppoed to the motion of the crate. a) hat i the work done by the weight? b) hat i the work done by the friction force? c) hat i the work done by Honoré? d) hat i the net work? cnx.org/content/m4150/latet/?collection=col11406/latet 3. The two following force act on an object F = i + j k N F = i + j + k N ( 4 ) ( 4 5 ) 1 hat i the net work done on the object if it i moving in a traight line from the poition (0 m, 1 m, m) to the poition (5 m, - m, - 3 m) then in a traight line once again from the poition (5 m, - m, - 3 m) to the poition (8 m, m, -5 m)? 4. Rita, whoe ma i 80 kg, ki downhill on a 30 lope. The coefficient of friction between the ki and the lope i 0.1. The ditance travelled by Rita i 300 m. vhcc.vhcc.edu/ph1fall9/frame_page/opentax_problem.htm a) hat i the work done by the weight between thee two intant? b) hat i the work done by the friction force between thee two intant? c) hat i the net work done on Rita between thee two intant? 019 Verion 8 ork 31

32 5. A pring initially compreed 50 cm puhe a 10 kg block. hat i the work done by the pring if the pring contant i 000 N/m? 6. An object i moving 6 m following the path hown in the figure. For the firt 3 m, a 40 N force toward the right act on the object. For the following m, an 80 N force toward the left act on the object. Finally, a 100 N force toward the right act on the object for the lat 1 m. hat i the work done on the object? 019 Verion 8 ork 3

33 7. Here the graph of the force acting on an object a a function of poition. hat i the work done on the object by thi force when the object move from x = 0 m to x = 6 m? 8. Here the graph of the force acting on an object a a function of poition. hat i the work done on the object by thi force when the object move from x = 4 m to x = 0 m? 019 Verion 8 ork 33

34 9. An object move along the path hown in the figure. Thi figure indicate the force exerted on the object on each traight part of the trajectory. hat i the work done on the object? 10. The force exerted on an object i given by the formula F x = x 18 N m ² hat i the work done on the object by thi force when the object move from x = 1 m to x = 3 m? 8. ork-energy Theorem 11. A 430 g occer ball i launched from the ground with an upward peed of 30 m/. Ue the work-energy theorem to find the peed of the ball when it i 0 m above it tarting point if the drag i neglected. 1. Mara, whoe ma i 5 kg, lide down the water lide hown in the figure. The coefficient of kinetic friction between Mara and lide i 0.1. Ue the work-energy theorem to determine the peed of Mara at the bottom of the lide Verion 8 ork 34

35 13. The frictional force doe 3000 J of work to completely top a kier who lide on a horizontal urface with an initial peed of 10 m/. vhcc.vhcc.edu/ph1fall9/frame_page/opentax_problem.htm hat would be the peed of the kier if the friction force had only done 1500 J of work? 14. René, whoe ma i 55 kg, tart a free-fall motion from a tationary air balloon. 10 econd after hi departure, René ha travelled 300 m, and hi peed i 39.4 m/. hat i the work done by the drag force during thee 10 econd? 15. A 00 g ma i attached to a pring whoe contant i 50 N/m. Originally, the pring i neither tretched nor compreed. Then the ma i releaed. hat will the maximum tretching of the pring be? en.wikipedia.org/wiki/hooke _law 019 Verion 8 ork 35

36 16. A 500 g ball i placed on a vertical pring whoe contant i 500 N/m. If the pring i compreed 10 cm from it equilibrium poition and then releaed, the ball reache a maximum height h max. hat i thi maximum height, meaured from the equilibrium poition of the pring? (Neglect the drag force.) 17. A 500 g block i placed in the pring gun hown in the figure. Initially, the pring i compreed 50 cm and then the block i releaed. There i no friction between the block and the lope. hat will the ditance travelled by the block be before it top? 18. Here the graph of the force acting on a 5 kg object a a function of it poition. quantumprogre.wordpre.com/011/01//reading-graph-with-your-lizard-brain/ The object ha a peed of m/ toward the poitive x-axi when it i at x = 0 m. a) hat i the peed of the object at x = 5 m? b) hat i the peed of the object at x = 1 m? 019 Verion 8 ork 36

37 19. A 10 kg block lide on a horizontal urface. The friction coefficient between the block and the urface i 0.1. hen the block i 1 m from a pring, whoe contant i 0 N/m, it ha a peed of m/. a) hat will the peed of the block be when the pring i compreed 0 cm? faraday.phyic.utoronto.ca/pvb/harrion/flah/tutorial/flahphyic.html b) hat will the maximum compreion of the pring be when it i hit by the object? faraday.phyic.utoronto.ca/pvb/harrion/flah/tutorial/flahphyic.html 8.3 Power 0. A BM 335i 007 i going at 10 km/h on a horizontal road. hat i the power of the engine of the car (in hp), knowing that the value of C x i 0.63 m² for thi car? (Auming that only the drag force i oppoed to the motion of the car and that the air denity i 1.3 kg/m³.) 019 Verion 8 ork 37

38 1. Laura lift a piano m above the ground in 0 econd uing the pulley ytem hown in the figure. hat i the average power of Laura (in hp)? (The piano ha no peed at the beginning and no peed at the end of thi motion.) A 5 kg crate i initially at ret. Then a winch pull it and give it a peed of 0 m/ in 10 econd. There i no friction between the crate and the urface. phyweb.bgu.ac.il/archive/coure/010a/physics1_indtmngmnt/010a/h5_q.php How much time will it take for a 100 kg crate initially at ret to reach a peed of 10 m/ if the ame winch (with the ame average power) i ued to pull the crate? 3. A winch pull a 50 kg box along a 30 lope with a contant peed of 10 m/. phyweb.bgu.ac.il/archive/coure/010a/physics1_indtmngmnt/010a/h5_q.php a) hat i the power of the winch if there i no friction between the box and the urface? b) hat i the power of the winch if the coefficient of kinetic friction between the box and the urface i 0.3? 019 Verion 8 ork 38

39 4. A 30 kg rocket toy lift off vertically from the ground. After 3 minute, it engine top. The rocket ha then a peed of 30 m/ and i at an altitude of 600 m. a) hat i the work done by the rocket engine? b) hat i the average power of the rocket engine? (Actually, the ma of the rocket decreae a fuel i ejected, but we will proceed a if the ma remain contant.) 5. A 0 kg object i ubjected to a force whoe power i given by P = 1 t Initially, the object i at ret. Thi i a one-dimenional motion along the x-axi. a) hat i the work done on the object between t = 0 and t = 10? b) hat i the peed of the object at t = 10? c) hat i the acceleration of the object at t = 10? d) hat i the force acting on the object at t = 10? e) hat i the diplacement of the object between t = 0 and t = 10? Challenge (Quetion more difficult than the exam quetion.) 6. The force acting on a 1 kg object initially at ret i given by the equation F = 9N 1 N m² x here doe the object reach it maximum peed and what i thi maximum peed? 019 Verion 8 ork 39

40 7. The force exerted on an object that can move in dimenion i given by the formula F = x + y i + y + x j ( 3 N ² 1 N ) ( N 1 N m m m m ) Thi object i moving from the point (1 m, 1 m) to the point ( m, m). a) hat i the work done on the object if it follow path A? b) hat i the work done on the object if it follow path B? c) hat i the work done on the object if it follow path C? 8.1 Definition of ork J. a) J b) J c) 35.9 J d) 750 J 3. 4 J 4. a) 117,600 J b) -0,369 J c) 97,31 J J J J J J J 8. ork-energy Theorem m/ m/ 019 Verion 8 ork 40

41 m/ ,010 J cm cm m 18. a) m/ b) The object cannot be at x = 1 cm 19. a) 1.5 m/ b) 63.5 cm 8.3 Power hp hp a) 450 b) a),300,400 J b) 1, a) 4000 J b) 0 m/ c) 3 m/² d) 60 N e) 80 m Challenge 6. 6 m/ at x = 3 m 7. a) 13 J b) 13 J c) 13 J 019 Verion 8 ork 41