13.4 Work done by Constant Forces

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1 13.4 Work done by Constnt Forces We will begin our discussion of the concept of work by nlyzing the motion of n object in one dimension cted on by constnt forces. Let s consider the following exmple: push cup forwrd with constnt force long desktop. When the cup chnges velocity (nd hence kinetic energy), the sum of the forces cting on the cup must be non-zero ccording to Newton s Second Lw. There re three forces involved in this motion: the pplied pushing force F ; the contct force C N + f k ; nd grvity F g = mg. The force digrm on the cup is shown in Figure Figure 13.2 Force digrm for cup. Let s choose our coordinte system so tht the +x -direction is the direction of the forwrd motion of the cup. The pushing force cn then be described by F = F x î. (13.4.1) Suppose body moves from n initil point x i to finl point x f so tht the displcement of the point the force cts on is Δx x f x i. The work done by constnt force F = F x î cting on the body is the product of the component of the force F x nd the displcement Δx, W = F x Δx. (13.4.2) Work is sclr quntity; it is not vector quntity. The SI unit for work is 13-1

2 [1 N m] = [1 kg m s ][1 m] = [1 kg m s ] = [1 J]. (13.4.3) Note tht work hs the sme dimension nd the sme SI unit s kinetic energy. Becuse our pplied force is long the direction of motion, both F x > 0 nd Δx > 0. In this exmple, the work done is just the product of the mgnitude of the pplied force nd the distnce through which tht force cts nd is positive. In the definition of work done by force, the force cn ct t ny point on the body. The displcement tht ppers in Eqution (13.4.2) is not the displcement of the body but the displcement of the point of ppliction of the force. For point-like objects, the displcement of the point of ppliction of the force is equl to the displcement of the body. However for n extended body, we need to focus on where the force cts nd whether or not tht point of ppliction undergoes ny displcement in the direction of the force s the following exmple illustrtes. Exmple 13.2 Work Done by Sttic Fiction Suppose you re initilly stnding nd you strt wlking by pushing ginst the ground with your feet nd your feet do not slip. Wht is the work done by the sttic friction force cting on you? Solution: When you pply contct force ginst the ground, the ground pplies n equl nd opposite contct force on you. The tngentil component of this constnt force is the force of sttic friction cting on you. Since your foot is t rest while you re pushing ginst the ground, there is no displcement of the point of ppliction of this sttic friction force. Therefore sttic friction does zero work on you while you re ccelerting. You my be surprised by this result but if you think bout energy trnsformtion, chemicl energy stored in your muscle cells is being trnsformed into kinetic energy of motion nd therml energy. When forces re opposing the motion, s in our exmple of pushing the cup, the kinetic friction force is given by! F f = f k,x î = µ k N î = µ mg î. (13.4.4) k Here the component of the force is in the opposite direction s the displcement. The work done by the kinetic friction force is negtive, W f = µ k mgδx. (13.4.5) Since the grvittion force is perpendiculr to the motion of the cup, the grvittionl force hs no component long the line of motion. Therefore the grvittion force does zero work on the cup when the cup is slid forwrd in the horizontl direction. The norml force is lso perpendiculr to the motion, nd hence does no work. 13-2

3 We see tht the pushing force does positive work, the kinetic friction force does negtive work, nd the grvittion nd norml force does zero work. Exmple 13.3 Work Done by Force Applied in the Direction of Displcement Push cup of mss 0.2 kg long horizontl tble with force of mgnitude 2.0 N for distnce of 0.5 m. The coefficient of friction between the tble nd the cup is µ k = Clculte the work done by the pushing force nd the work done by the friction force. Solution: The work done by the pushing force is The work done by the friction force is W = F x Δx = (2.0 N)(0.5 m) = 1.0 J. (13.4.6) W f = µ k mgδx = (0.1)(0.2 kg)(9.8 m s -2 )(0.5 m)= 0.10 J. (13.4.7) Exmple 13.4 Work Done by Force Applied t n Angle to the Direction of Displcement Suppose we push the cup in the previous exmple with force of the sme mgnitude but t n ngle θ = 30 o upwrds with respect to the tble. Clculte the work done by the pushing force. Clculte the work done by the kinetic friction force. Solution: The force digrm on the cup nd coordinte system is shown in Figure Figure 13.3 Force digrm on cup. The x -component of the pushing force is now The work done by the pushing force is F x = F cos(θ ) = (2.0 N)(cos(30 )) = 1.7 N. (13.4.8) 13-3

4 W = F x Δx = (1.7 N)(0.5 m) = J. (13.4.9) The kinetic friction force is F f = µ k N î. ( ) In this cse, the mgnitude of the norml force is not simply the sme s the weight of the cup. We need to find the y -component of the pplied force, F y = F sin(θ ) = (2.0 N)(sin(30 o ) = 1.0 N. ( ) To find the norml force, we pply Newton s Second Lw in the y -direction, Then the norml force is F y + N mg = 0. ( ) N = mg F y = (0.2 kg)(9.8 m s 2 ) (1.0 N) = N. ( ) The work done by the kinetic friction force is W f = µ k N Δx = (0.1)( N)(0.5 m) = J. ( ) Exmple 13.5 Work done by Grvity Ner the Surfce of the Erth Consider point-like body of mss m ner the surfce of the erth flling directly towrds the center of the erth. The grvittion force between the body nd the erth is nerly constnt, F grv = mg. Let s choose coordinte system with the origin t the surfce of the erth nd the + y -direction pointing wy from the center of the erth Suppose the body strts from n initil point y i nd flls to finl point y f closer to the erth. How much work does the grvittion force do on the body s it flls? Solution: The displcement of the body is negtive, Δy y f y i < 0. The grvittion force is given by F g g = mg = F y ĵ = mg ĵ. ( ) The work done on the body is then W g g = F y Δy = mgδy. ( ) 13-4

5 For flling body, the displcement of the body is negtive, Δy y f y i < 0 ; therefore the work done by grvity is positive, W g > 0. The grvittion force is pointing in the sme direction s the displcement of the flling object so the work should be positive. When n object is rising while under the influence of grvittion force, Δy y f y i > 0. The work done by the grvittion force for rising body is negtive, W g < 0, becuse the grvittion force is pointing in the opposite direction from tht in which the object is displced. It s importnt to note tht the choice of the positive direction s being wy from the center of the erth ( up ) does not mke difference. If the downwrd direction were chosen positive, the flling body would hve positive displcement nd the grvittionl force s given in Eqution ( ) would hve positive downwrd component; the product F y g Δy would still be positive Work done by Non-Constnt Forces Consider body moving in the x -direction under the influence of non-constnt force in the x -direction, F = F x î. The body moves from n initil position x i to finl position x f. In order to clculte the work done by non-constnt force, we will divide up the displcement of the point of ppliction of the force into lrge number N of smll displcements Δx j th j where the index j mrks the displcement nd tkes integer vlues from 1 to N. Let (F x, j ) ve denote the verge vlue of the x -component of the force in the displcement intervl [x j 1, x j ]. For the clculte the contribution to the work j th displcement intervl we W j = (F ) Δx j (13.5.1) x, j ve This contribution is sclr so we dd up these sclr quntities to get the totl work j=n j=n W N = W j = (F ) Δx j. (13.5.2) x, j ve j=1 j=1 The sum in Eqution (13.5.2) depends on the number of divisions N nd the width of the intervls Δx j. In order to define quntity tht is independent of the divisions, we tke the limit s N nd Δx j 0 for ll j. The work is then 13-5

6 j= N x=x f W = lim (F x, j ) ve Δx j = F x N j=1 Δx x=x i j 0 (x) dx (13.5.3) This lst expression is the definite integrl of the x -component of the force with respect to the prmeter x. In Figure 13.5 we grph the x -component of the force s function of the prmeter x. The work integrl is the re under this curve between x = x i nd x = x f. Figure 13.5 Plot of x -component of smple force F x (x) s function of x. Exmple 13.6 Work done by the Spring Force Connect one end of n unstretched spring of length l 0 with spring constnt k to n object resting on smooth frictionless tble nd fix the other end of the spring to wll. Choose n origin s shown in the figure. Stretch the spring by n mount x i nd relese the object. How much work does the spring do on the object when the spring is stretched by n mount x f? l 0 x i î l 0 î l 0 î x f x = 0 x = 0 x = 0 Figure 13.6 Equilibrium, initil nd finl sttes for spring Solution: We first begin by choosing coordinte system with our origin locted t the position of the object when the spring is unstretched (or uncompressed). We choose the î unit vector to point in the direction the object moves when the spring is being stretched. We choose the coordinte function x to denote the position of the object with respect to the origin. We show the coordinte function nd free-body force digrm in the figure below. 13-6

7 l 0 x î x F = F x î = kx î î x = 0 x = 0 Figure 13.6 Spring force The spring force on the object is given by (Figure 13.6)! F = F x î = k x î (13.5.4) In Figure 13.7 we show the grph of the x -component of the spring force, F x (x), s function of x. F x (x) x f x i +x F x (x) = k x Figure 13.7 Plot of spring force F x (x) vs. displcement x The work done is just the re under the curve for the intervl x i to x f, x =x f x =x f W = F ( x = x kx dx = 1 2 k(x 2 x )d x 2 ) (13.5.5) f i x =x i x =x i This result is independent of the sign of x i nd x f becuse both quntities pper s squres. If the spring is less stretched or compressed in the finl stte thn in the initil stte, then the bsolute vlue, x f < x i, nd the work done by the spring force is positive. The spring force does positive work on the body when the spring goes from stte of greter tension to stte of lesser tension. 13-7

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