4.1 Work Done by a Constant orce work the prodct of the magnitde of an object s and the component of the applied force in the direction of the Stdying can feel like a lot of work. Imagine stdying several hors for a difficlt test or spending all afternoon writing a report for class. While this is a significant amont of hard work, in the scientific sense of the word, yo have done no work at all. In physics, work is the energy that a force gives to an object when the force moves the object. When yo read yor notes, yo do no work becase yo do not exert a force on an object to move it. Yor everyday life, however, is filled with examples of work in the scientific sense. Yo do work on a backpack when yo lift it to yor sholders. Yo do work on the classroom door when yo psh it open or pll it closed. Yo do work on a basketball by boncing or throwing it. In these cases, a force does work on an object to move it. Yo may notice that yo do more work lifting a backpack filled with heavy books than lifting an empty one, and more work lifting a backpack from the floor than from a desk becase the distance from the floor to yor sholder is greater than the distance from a desk to yor sholder. The scientific definition of work is consistent with these experiences. Work Work depends on the magnitde of the force applied to an object and the distance the object moves. In fact, work depends only on the object s in the direction of the applied force and not perpendiclar to the force. In other words, the work depends only on the component of force in the direction of motion and not the force perpendiclar to the motion. Sppose, for example, yo are shovelling snow. Yo might psh on the snow at an angle θ to the grond, bt the snow moves horizontally (igre 1). The component of force directed into the grond does not help move the snow across the grond. Only the component of force along the grond moves the snow. In any sitation, only the force in the direction of an object s does work on the object. The eqation for calclating the work, W, that a constant force, >, does to case the, Dd >, of an object is W 5 d cos θ igre 1 The force of the shovel does work on the snow. Only the horizontal force component contribtes to the work. where is the magnitde of the force, d is the magnitde of the object s, and θ is the angle between the force and the. The component of the force in the direction of the motion is cos θ (igre 2). sin cos y (a) direction igre 2 (a) A force acts at an angle to the. (b) Only the component parallel to the does work. (b) d x 164 Chapter 4 Work and Energy NEL
The work done by a force depends on two vectors: the applied force and the reslting. The work done on an object is proportional to both force, >, and, Dd >. However, the amont of work done is a scalar qantity, not a vector. The SI nits of work are newton-metres (N?m), or kilograms times metres sqared per second sqared (kg?m 2 /s 2 ). This nit is called a jole (J). In the following Ttorial, yo will explore how to se the work eqation to determine the work done in different sitations. jole the SI nit of work and energy; a force of 1 N acting over a of 1 m does 1 J of work; symbol J Ttorial 1 Calclating Work Done Yo can determine the amont of work done on an object if yo know the applied force, the of the object, and the angle between the force and the. Sample Problem 1: Calclating the Work Done When the orce Is Parallel to the Displacement The of an object is often in the same direction as the applied force. In this case, the eqation for calclating the work done by the force is simplifi ed. Sppose a hockey player slides a pck along the ice with a constant force of 85 N in the forward direction (igre 3). The pck moves a horizontal distance of 0.20 m while in contact with the hockey stick. Calclate the amont of work done on the pck by the stick. = 85 N Dd = 0.20 m Given: 5 85 N; d 5 0.20 m Analysis: Use the eqation for work, W 5 d cos. and d are in the same direction, so the angle between them is zero, 5 0. Soltion: W 5 d cos 5 (85 N)(0.20 m) cos 08 5 (17 N?m)(1) 5 17 N?m W 5 17 J Statement: The hockey stick does 17 J of work on the pck. Note that the stick only does work on the pck while the stick is in contact with the pck, applying force. igre 3 Sample Problem 2: Calclating the Work Done When the orce Is at an Angle to the Displacement The applied force and the can be at an angle to each other. or example, a stdent pshes a lawnmower forward with a constant force of 48 N for a distance of 7.5 m (igre 4). The angle between the force and the of the lawnmower is 32. Calclate how mch work is done on the lawnmower by the stdent. sin d y x Given: 5 48 N; d 5 7.5 m; 5 328 Analysis: The work done on the lawnmower by the stdent depends only on the component of force in the direction of the mower s. Use the eqation for work, W 5 d cos. and d are at an angle of 328 to each other. Soltion: W 5 d cos 5 (48 N)(7.5 m) cos 328 5 (360 N?m)(0.848) W 5 3.1 3 10 2 J Statement: The stdent does 3.1 3 10 2 J of work on the lawnmower. igre 4 direction cos NEL 4.1 Work Done by a Constant orce 165
Sample Problem 3: Calclating the Work Done When the orce Is Perpendiclar to the Displacement Sppose yo carry books with a weight of 22 N a distance of 3.8 m across a room (igre 5). Determine the work done on the books. direction igre 5 Given: 5 22 N; d 5 3.8 m; 5 908 Analysis: By Newton s third law, the pward force yo exert on the books is eqal to the weight of the books, 22 N. Use the eqation for work, W 5 d cos. and d are at an angle of 908 to each other. Soltion: W 5 d cos 5 (22 N)(3.8 m) cos 908 5 (22 N)(3.8 m)(0) 5 0 N?m W 5 0 J Statement: No work is done on the books. 1. A weightlifter ses a force of 275 N to lift weights directly pward throgh a distance of 0.65 m. Determine the work done on the weights by the weightlifter. T/I A [ans: 1.8 3 10 2 J] 2. Calclate the work done on a wall if yo psh on it with a constant force of 9.4 N and the wall does not move. T/I A [ans: 0 J] 3. A pool ce stick strikes a ball with a constant force of 0.73 N, casing the ball to move 0.65 m in the direction of the force. The ball moves 0.080 m while the ce stick is in contact with it. Calclate the work done on the ball by the ce stick. T/I A [ans: 0.058 J] 4. A tow trck ses a winch with a rope attached to pll a car that is stck in a ditch. The rope exerts a force of 9.9 3 10 3 N on the car body, and the angle between the rope and the direction the car moves is 12. Determine the amont of work done on the car by the tow trck to move the car 4.3 m. T/I A [ans: 4.2 3 10 4 J] Positive and negative work The force of the hockey stick on the pck in Sample Problem 1 reslted in positive work that cased the pck s speed to increase. What if the pck slides to rest? The of the pck and the force on the pck by friction are in opposite directions. When an object moves in a direction opposite to an applied force, the force does negative work. Negative work will case a loss of kinetic energy. orces that case negative work are exerted at an angle between 908 and 1808, opposite to the object s direction. As yo read in Chapter 2, friction is the force resisting the motion of objects moving against each other. Ttorial 2 gives yo an opportnity to calclate the negative work that occrs when a force, sch as friction, and are in opposite directions. Ttorial 2 Calclating Negative Work Sample Problem 1: Calclating Negative Work in One Dimension Sppose a car is moving along a straight road when the driver sddenly applies the brakes. The force of friction between the grond and the car tires is opposite to the car s direction of motion and decreases the car s speed. Calclate the work done by a constant frictional force of 1.4 kn over a distance of 1.2 3 10 2 m. Given: 5 1.4 kn 5 1.4 3 10 3 N; Dd 5 1.2 3 10 2 m; 5 1808 Analysis: W 5 Dd cos θ 166 Chapter 4 Work and Energy NEL
Soltion: W 5 Dd cos 5 (1.4 3 10 3 N)(1.2 3 10 2 m) cos 1808 5 (1.4 3 10 3 N)(1.2 3 10 2 m)(21) 5 21.70 3 10 5 N?m W 5 21.70 3 10 5 J Statement: riction does 21.7 3 10 5 J of work on the car to slow it down. The negative sign indicates that the force is opposing the motion. Sample Problem 2: Calclating Negative Work in Two Dimensions An ice skater slides to a stop by pshing her blades against the ice (igre 6). The ice exerts a constant force of 95 N on the skater, and the skater stops in 1.2 m. The angle between the force and the skater s direction of motion is 1408. Calclate the work done on the skater by the ice. 95 N 140 1.2 m igre 6 Given: 5 95 N; Dd 5 1.2 m; 5 1408 Analysis: Use the work eqation, W 5 Dd cos. Remember that cos is negative for angles between 908 and 1808. Soltion: W 5 Dd cos 5 (95 N)(1.2 m) cos 1408 5 287 N?m W 5 287 J Statement: The ice does 87 J of work on the skater. 1. A drop tower ride lifts riders at a constant speed to a height of 78 m and sddenly drops them. T/I A (a) Determine the work done on a 56 kg rider by the machine as she is lifted to the top of the ride. [ans: 4.3 3 10 4 J] (b) Determine the work done on the rider by gravity as she is lifted to the top of the ride. [ans: 24.3 3 10 4 J] 2. As a passenger airplane toches down it skids across the rnway to a stop. riction between the grond and the plane s wheels applies a constant force of 5.21 kn as the plane slides a distance of 355 m. T/I A (a) Calclate the work done on the airplane by friction. [ans: 21.85 3 10 6 J] (b) Determine the distance the plane wold slide if friction applied the same force bt did 21.52 3 10 6 J of work. [ans: 292 m] 3. A skier slides down a snowy hill and then stops by pressing his skis at an angle to the snow. The snow exerts a constant force of 5.9 N on the skier at an angle of 1508 to the skier s. The skier moves a distance of 3.5 m. Calclate the work done on the skier by the snow. [ans: 218 J] Yo observed in Ttorial 1 that a force does positive work on an object when the object s is in the same direction as the force. The work is also positive when the direction of the object s is at an angle between 08 and 908 to the applied force. Similarly, Ttorial 2 showed that a force does negative work on an object when the object s is opposite to the direction of the force, at an angle between 908 and 1808 to the applied force. Recall that zero work is done on an object when the object s is exactly 908 to the applied force. Are there other sitations involving an applied force in which zero work is done? zero work When the direction of an object s is exactly 908 to the applied force, the force does no work on the object. Look back at Sample Problem 3 in Ttorial 1. As the stdent walks across the room, he pshes p on the books, bt the books move to the right. The angle between the force and the is 908. Since W 5 Δd cos θ and cos 908 5 0, the stdent does no work on the books. The work also eqals zero when the force on an object is zero. Consider a probe travelling in space far from any gravitational forces. The probe s motion is de to inertia, bt the force on it is zero. As a reslt, zero work is done on the probe. NEL 4.1 Work Done by a Constant orce 167
The third variable in the work eqation is the of the object. Have yo ever tried to twist the lid off a jar, bt the lid was stck? Regardless of how mch force yo exert on the lid, yo do zero work nless the lid moves. The work on an object is zero when any of the force, the, or the cosine of the angle between the force and the is zero. In the following Ttorial, we will examine another example of a force that does zero work. Ttorial 3 Work and Centripetal Acceleration When a centripetal force acts on an object moving along a circlar path, the direction of the force is perpendiclar to the direction of the object s motion. Sample Problem 1: Calclating the Work Done on an Object Moving in a Circlar Path A stdent attaches one end of a string to a rbber stopper. She then holds the other end of the string and twirls the rbber stopper in a horizontal circle arond her head. The string exerts a tension force on the stopper directed toward the centre of the circle (igre 7). Determine the work done on the stopper by the string dring a revoltion. direction of stopper s motion rbber stopper T force over a short time interval direction of over a short time interval path of stopper (a) (b) igre 7 (a) The circlar path of the rbber stopper, seen from above. (b) Dring a very short time interval, the of the stopper is perpendiclar to the tension force. T v Given: 5 908 Analysis: The tension force, > T, cases the stopper s centripetal acceleration. At each moment, the stopper s instanta neos velocity is at an angle of 908 to the tension force. Dring a very short time interval, the very small of the stopper is also at an angle of 908 to the tension force. We can break one loop arond into a series of many small s, each occrring dring a very short time interval. Dring each time interval, the tension and the are perpendiclar. The total work done dring one loop arond will eqal the sm of work done dring each small. or each small, se the work eqation, W 5 Dd cos, with 5 908. Soltion: W 5 Dd cos 5 Dd cos 908 5 Dd (0) W 5 0 J Statement: Smming the work done dring all the small s of the loop gives a total of W 5 0 J dring each revoltion. The tension force exerted by the string does zero work on the stopper dring the revoltion. 1. Earth exerts a gravitational pll that cases the Moon to experience a centripetal acceleration dring its orbit. Assme that the Moon s orbit arond Earth is circlar. Determine the work done by Earth s gravitational pll on the Moon. As part of yor soltion, inclde a diagram that illstrates Earth, the Moon, the direction of travel, and the force at one instant in time. K/U T/I C [ans: 0 J] In Ttorial 3, the work does not depend on the magnitde of the tension force or the total distance the stopper moves. We treated the stopper s orbit as a series of tiny s perpendiclar to the tension force. or each tiny, the tension force does zero work on the stopper. We conclde that the work done by the centripetal force acting on an object in circlar motion is zero. 168 Chapter 4 Work and Energy NEL
Work Done by Mltiple orces Almost all real-world examples of work involve friction pls other forces. Ttorial 4 explores how the presence of mltiple forces affects the work done on an object. Ttorial 4 Calclating Work Done by Mltiple orces Sample Problem 1: Calclating the Work Done by Mltiple orces on a Dragged Object igre 8 shows a long-distance hiker plling a sled across a snowy field. f y N h g h x igre 8 The hiker exerts a constant force of 135 N on the sled at a 48.08 angle to the sled s. At the same time, a constant 67.0 N frictional force on the sled from the snow opposes the motion. The sled also experiences the force from gravity and the normal force from the snow, bt these forces do not contribte to the work. Calclate the work done by the hiker (h), the work done by friction (f), and the total (T) work done on the sled when the hiker plls the sled 345 m over the snow. Given: h 5 135 N; f 5 67.0 N; Dd 5 345 m; h 5 48.08; f 5 1808 Analysis: W 5 Dd cos. The total work done is the sm of the work done by the individal forces. Soltion: W h 5 h Dd cos h 5 (135 N)(345 m) cos 48.08 5 3.116 3 10 4 N?m W h 5 3.116 3 10 4 J (one extra digit carried) W f 5 f Dd cos f 5 (67.0 N)(345 m) cos 1808 5 22.312 3 10 4 N?m W f 5 22.312 3 10 4 J (one extra digit carried) W T 5 W h 1 W f 5 3.116 3 10 4 J 2 2.312 3 10 4 J W T 5 8.04 3 10 3 J Statement: The hiker does 3.12 3 10 4 J of work on the sled. riction does 22.31 3 10 4 J of work on the sled. The total work done on the sled is 8.04 3 10 3 J. h ; W f ; W T 1. A hiker plls a sled a distance of 223 m with a constant force of 122 N exerted at an angle of 378. riction acts on the sled with a constant force of 72.3 N. Calclate the work done on the sled by the hiker and by friction, and the total work done on the sled. T/I A [ans: 2.2 3 10 4 J; 21.6 3 10 4 J; 5.6 3 10 3 J] 2. If together the hiker and friction do 2.42 3 10 4 J of total work on the sled in igre 8, how far did the hiker pll the sled? T/I A [ans: 963 m] A frictional force is present in many parts of the hiker and sled problem in Ttorial 4. riction acts, for example, between the hiker s hands and the rope, between the rope and the sled, and between the hiker s boots and the snow. A frictional force acts and does work on any srfaces that slide past each other. In each case, the frictional force transfers energy to srfaces, increasing their kinetic energy. The energy does not disappear. The increase in temperatre occrs becase of the motion of atoms at the srfaces. All sbstances are composed of particles in constant motion. When friction heats two srfaces, the vibrations of the particles become larger, and the particles move more qickly. The faster-moving particles have more energy, and the energy stored in the chemical bonds increases. In smmary, when analyzing the total work done on an object, all forces that are present, inclding friction, mst be considered. The net effect of these forces can reslt in either positive, negative, or zero total work done on the object. NEL 4.1 Work Done by a Constant orce 169
4.1 Review Smmary Work occrs when a force > is applied to move an object a Dd >. The work done on an object is given by W 5 Dd cos, where is the angle between > and d >. The SI nit of work is the jole (J). When an object moves at an angle to an applied force, only the component of the force in the direction of the does work on the object. When an object moves in a direction opposite to an applied force, the force does negative work on the object. A force does zero work on an object when the angle between the force and the object s is 908. Zero work is also done when either the force or the is zero. Qestions 1. Two ropes pll on a crate toward the left with forces of eqal magnitde,, casing the crate to move horizontally (igre 9). Which rope does more work on the crate? Explain. K/U 5. A warehose worker pshes a crate of mass 24 kg p a ramp (igre 11). Assme that the friction between the crate and ramp can be ignored. K/U T/I 2 1 5 30.0 igre 9 2. A toy consists of a small plastic tbe connected at the centre to one end of a long string. A girl holds the other end of the string and swings the toy in a horizontal, circlar path above her head. Is work done on the toy by the string dring each revoltion? Explain yor reasoning. K/U 3. A shopper pshes a loaded grocery cart with a force of 12.6 N. The force makes an angle of 21.88 above the horizontal. Determine the work done on the cart by the shopper as he pshes the cart 14.2 m. T/I A 4. The horse in igre 10 plls a rider on a sleigh across a snowy horizontal field. The force of the rope is 22.8 N, and the horse does 9.53 3 10 2 J of work plling the sleigh a distance of 52.6 m. Calclate the angle between the rope and the horizontal. T/I A igre 11 (a) Determine the component of gravitational force directed along the ramp s srface. (b) Calclate the force reqired to move the crate at a constant speed p the ramp. (c) Calclate the work done in pshing the crate 23 m as measred along the ramp. Assme the crate moves at a constant velocity. (d) Assme the coefficient of kinetic friction between the crate and the ramp is μ K 5 0.25. Calclate the work done on the crate by the worker and by friction, and calclate the total work done as the worker pshes the crate 16 m p the ramp. 6. A boy and a girl pll and psh a crate along an icy horizontal srface, moving it 13 m at a constant speed. The boy exerts 75 N of force at an angle of 328 above the horizontal, and the girl exerts a force of 75 N at an angle of 228 above the horizontal. Calclate the total work done by the boy and girl together. T/I igre 10 170 Chapter 4 Work and Energy NEL