University Physics Al No 6 Work, Energy, and the Cwe Theorem Class Number ame I. Choose the Correct Answer 1. Which of the following quantities are independent of the choice of inertial frame?(There may be more than one correct answer) (BC) (A)Velocity B)Acceleration (C)Force (D)Work In all the inertial frame, the forces acted on a body are same F= ma=F=ma, therefore the force and acceleration are independent of the choice of inertial frame 2. The force exerted by a special compress device is given by F(x)=lx(x-1)for 0<xsl where / is the maximum possible compression and k is a constant. The force required to compress the device a distance d is a maximum when (A)d=0.(B)d=l/4.(C)d=l/√2.(D)d=1/2.(E)d=l Solution The force is F(x)=k(x=)=k(sI2 k/. when 0 or x1, The force is a maximum According to the problem, x=d-l 3. A particle has a constant kinetic energy KE. Which of the following quantities must also be constant? (A)Position (B)Speed (C)Velocity D)Momentum Solution Since the kinetic energy is KE=-my2, it is dependent of the speed 4. A020kg puck slides across a frictionless floor with a speed of 10 m/s. The net work done on the (B)-10J (C)0J (D)20J Solution Using the CWE theorem: otal=AKE, the net work done on the puck is zero
University Physics AI No. 6 Work, Energy, and the CWE Theorem Class Number Name I．Choose the Correct Answer 1. Which of the following quantities are independent of the choice of inertial frame? (There may be more than one correct answer) （ BC ） (A) Velocity (B) Acceleration (C) Force (D) Work Solution： In all the inertial frame, the forces acted on a body are same, F = ma = F′ = ma′ r r r r , therefore the force and acceleration are independent of the choice of inertial frame. 2. The force exerted by a special compress device is given by F (x) kx(x l) x = − for 0 ≤ x ≤ l , where l is the maximum possible compression and k is a constant. The force required to compress the device a distance d is a maximum when （ E ） (A) d = 0 . (B) d = l / 4 . (C) d = l / 2 . (D) d = l / 2 . (E) d = l Solution: The force is 4 ) 2 ( ) ( ) ( 2 l 2 kl F x kx x l k x x = − = − − , when x=0 or x=l, The force is a maximum. According to the problem, x=d=l 3. A particle has a constant kinetic energy KE. Which of the following quantities must also be constant? （ B ） (A) Position (B) Speed (C) Velocity (D) Momentum Solution: Since the kinetic energy is 2 2 1 KE = mv , it is dependent of the speed. 4. A 0.20kg puck slides across a frictionless floor with a speed of 10 m/s. The net work done on the puck is （ C ） (A) -20J (B) -10J (C) 0J (D) 20J Solution： Using the CWE theorem: Wtotal = ∆KE , the net work done on the puck is zero
5. Which of the following forces is not conservative? (A)F=3+4)(B)F=3xi+4y(C)F=3y+4x(D)F=3x2+4y2 Solutio The work done by the force in(C)is Fdi=3ydr+]4xdy, the result of this integration depends on the rout of the path I. filling the blanks 1. A chain is held on a frictionless table with one-fourth FEEEEEEEEEEEERE of its length hanging over the edge, as shown in Fig. 1. If the chain has a length L and a mass m, the work required to pull the hanging part back on the table is mgL/32 Solution Since Conserv =-APE, if we choose the zero potential energy on the table the work required to pull the hanging part back on the table is W gh L L a forc system has only -component Fr; a graph of this compone FN 30 Irc function of position gure 2. The work done by this force if the 20 system moves along the x-axis from x=0 m to x= 10.0 m is 235.5 J x(m) 100 The work done by this force is the Fig 2 curve between x=0 m and 丌(30N)(5m)=75=235.5→W=235.5J 3. A 2. 14kg block is dropped from a height of 43 6cm onto a spring of force constant k=186N/cm, as shown in Fig 3. The maximum 43.6cm ill be Solution Using conversation of energy, assume the gravitational potential Fig 3
5. Which of the following forces is not conservative? （ C ） (A) F i j = 3ˆ + 4 ˆ r (B) F xi yj = 3 ˆ + 4 ˆ r (C) F yi xj = 3 ˆ + 4 ˆ r (D) F x i y j 3 ˆ 4 ˆ 2 2 = + r Solution The work done by the force in (C) is ∫ ∫ ∫ ⋅ = + f i f i x x y y F dr 3ydx 4xdy v v , the result of this integration depends on the rout of the path. II. Filling the Blanks 1. A chain is held on a frictionless table with one-fourth of its length hanging over the edge, as shown in Fig.1. If the chain has a length L and a mass m, the work required to pull the hanging part back on the table is mgL/32 . Solution: Since Wconserv = −∆PE , if we choose the zero potential energy on the table the work required to pull the hanging part back on the table is W mgh mg L mgL 32 1 8 1 4 1 conserv = = ⋅ = 2 A force on a system has only an x-component Fx; a graph of this component as a function of position x is shown in Figure 2. The work done by this force if the system moves along the x-axis from x = 0 m to x = 10.0 m is 235.5 J . Solution: The work done by this force is the area under the curve between xi = 0 m and x f =10 m (30N)(5m) 75 235.5 235.5 J 2 1 area = π = π = ⇒ W = 3. A 2.14kg block is dropped from a height of 43.6cm onto a spring of force constant k=18.6N/cm, as shown in Fig.3. The maximum distance the spring will be compressed is 11.1cm . Solution: Using conversation of energy, assume the gravitational potential x(m) Fx(N) Fig.2 5.0 10.0 10 20 30 Circle Fig.1 k 43.6cm Fig.3
energy is zero at the initial position. So we have 0=-mg(43.6+x)+k →x2-2.255x-9832=0→x=11.1 4. The work done by the gravitational force in moving a system of mass m at a distance r from the center of a uniform sphere(of mass GMm M) to a distance reis ) where both ri and rg R e less than the radius R of the sphere(see Figure 4). This imply about the work done by this form of the gravitational force around a closed path is zero Solution Since the mass is in the uniform sphere, te gravitational force is a function of position r, so the gravitational force is GmM 4 FG △ →W=dF=( GMm r)·drr=- R 5. a particle moves along the x axis under the influence of a conservative force that is described by F=-axe where a and are constants. The potential energy function U(x)is 2B Solution According to the definition of the potential energy, we ge U(x)=[.d=-areeaxdxsareaxar 3peaai0eva2 6. A force F=-kxi-hy is one of the forces on a particle of mass m. The quantity k is constant This force represents a two-dimensional Hooke's force law. Is this force conservative?yes The appropriate potential energy function for this force is -k(x+y)
energy is zero at the initial position. So we have 2.255 98.32 0 11.1 2 1 0 (43.6 ) 2 2 ⇒ − − = ⇒ = = − + + x x x mg x kx 4. The work done by the gravitational force in moving a system of mass m at a distance ri from the center of a uniform sphere (of mass M) to a distance rf is ) 2 2 ( 2 2 3 i rf r R GMm − − , where both ri and rf are less than the radius R of the sphere (see Figure 4). This imply about the work done by this form of the gravitational force around a closed path is zero . Solution: Since the mass is in the uniform sphere, te gravitational force is a function of position r, so the gravitational force is rr R GMm r r R M r Gm FG ˆ ˆ 3 4 3 4 3 3 3 2 = − ⋅ ⋅ π = − ⋅ π r ) 2 2 ( 2 d ( )ˆ dr 2 2 3 2 3 3 | i f r r r r f i G r r R r GMm R GMm r r r R GMm W F r f i f i ⇒ = ⋅ = − ⋅ ⋅ = − ⋅ = − − ∫ ∫ r r r 5. A particle moves along the x axis under the influence of a conservative force that is described by F xe i x ˆ 2 β α − = − r , where α and β are constants. The potential energy function U(x) is 2 2 x e β β α − . Solution: According to the definition of the potential energy, we get 2 2 2 2 2 2 d 2 ( ) d d 2 x x x x x x x p U x F r xe x e x e e β β β β β α β α α α − ∞ − ∞ − ∞ − ∞ = ⋅ = − = − = − = ∫ ∫ ∫ v r 6. A force F kxi kyj = − ˆ − ˆ r is one of the forces on a particle of mass m. The quantity k is constant. This force represents a two-dimensional Hooke’s force law. Is this force conservative? yes . The appropriate potential energy function for this force is ( ) 2 1 2 2 k x + y . m R Fig.4 ir r f r r
Solution (a)Assume that the particle moves from the position (x i+y,j+= k) to the position r,(x i+y,j Thus the work done by the force F=-hi-kyj is dr=l(-kxi )dxi+'(kyj )dy 2)--k(y2-y2) Notice that the work done by the force depends only on the initial and final position. So the force is conservative W=-[k(x2+y2)--k( D]=-(PE,-PEi) Potential energy function for this force is PE=k(x+y) II. Give the solutions of the following problems An 80.0 kg sphere is suspended by a wire of length 25.0 m from the ceiling of a science museum as indicated in Figure 5. A horizontal force of magnitude F is applied to the ball, moving it very slowly at B constant speed until the wire makes an angle with vertical direction 35° (a)Sketch a second law force diagram indicating all the forces on the ball at any point along the path (b) Is the force needed to accomplish the task constant in magnitude 0 along the path followed by the ball? (c)Use the CWE theorem to find the work done by the force F Solution: (a)Set up the coordinate system, and sketch the second law force diagram is shown in figure (b) Apply Newtons law to the sphere ∫F-7sinb=0 Img-Tcos0=0 Since F=mg tan 8 and 0 is changed from 0 to 35, the force F is not constant (c)Let the origin be zero potential energy, apply the Cwe theorem WF=△E=-mg(lcos35°-1)=-80×981×25×(.82-1)=354824(J)
Solution: (a) Assume that the particle moves from the position ) ˆ ˆ ˆ r (x i y j z k i i + i + i r to the position ) ˆ ˆ ˆ r (x i y j z k f f + f + f r . Thus the work done by the force F kxi kyj = − ˆ − ˆ r is ∫ ∫ ∫ = ⋅ = − + − = − − − − f i f i f i x x y y f i f i r r W F r kxi kyj k x x k( y y ) 2 1 ( ) 2 1 j ˆ i ( ˆ)dy ˆ d ( ˆ)dx r 2 2 2 2 r Notice that the work done by the force depends only on the initial and final position. So the force is conservative. (b) ( )] ( ) 2 1 ( ) 2 1 [ 2 2 2 2 f f i i PEf PEi W = − k x + y − k x + y = − − Potential energy function for this force is ( ) 2 1 2 2 PE = k x + y III. Give the Solutions of the Following Problems 1. An 80.0 kg sphere is suspended by a wire of length 25.0 m from the ceiling of a science museum as indicated in Figure 5. A horizontal force of magnitude F is applied to the ball, moving it very slowly at constant speed until the wire makes an angle with vertical direction equal to 35°. (a) Sketch a second law force diagram indicating all the forces on the ball at any point along the path. (b) Is the force needed to accomplish the task constant in magnitude along the path followed by the ball? (c) Use the CWE theorem to find the work done by the force F r . Solution: (a) Set up the coordinate system, and sketch the second law force diagram is shown in figure. (b) Apply Newton’s law to the sphere ⎩ ⎨ ⎧ − = − = cos 0 sin 0 θ θ mg T F T Since F = mg tanθ and θ is changed from 0° to 35°, the force F is not constant. (c) Let the origin be zero potential energy, apply the CWE theorem W = ∆E = −mg(l cos35 − l) = −80×9.81× 25× (0.82 −1) = 3548.24(J) F o F r 35° y x 0 T v mg v
2. A particle of mass 2. Okg moves along the x axis through a region in which its potential energy U(x) varies as shown in Fig. 6. When the particle is at -5F-=-- x=2.0m, its velocity is -20m/s (a)Calculate the force acting on the particle at this position. (b)Between.10 what limits does the motion take place?(c)How fast is s it moving when it is at x7. 0m? Solution (a) Using the relationship of F =-,(PE),from Fig6 the diagram, the force acting on the particle at x=2.0m the opposite of the tangent of the line L12. So F (PB)≈17-3 4.7N (b) At the point x20m, we can calculate the total mechanical energy E=PE+Ke=-8 As we must have E> PE, PE s-4J Look at the diagram, the position limit of the particle is 1.5m≤x≤141 (c)E=PE+KE→-17+×2xy2=-4→w=√13(m/s) 3. A projectile is launched at initial speed vo at an angle 0 with the horizontal direction. Neglect air friction and assume the trajectory is confined to a region near the surface of the Earth. Find the maximum vertical height of the projectile at the top of its flight path by using CWe theorem Solution Set up the coordinate system shown in figure, choose the elevation of the y -coordinate origin The initial kinetic energy KE=mvo The initial potential energy is PE=mgy=0 The kinetic energy at the top of its flight path is KEop=m(vo cos 0) The potential energy at the top of its flight path is PErop=mgy op Making the appropriate substitution into the Cwe theorem
2. A particle of mass 2.0kg moves along the x axis through a region in which its potential energy U(x) varies as shown in Fig. 6. When the particle is at x=2.0m, its velocity is -2.0m/s. (a) Calculate the force acting on the particle at this position. (b) Between what limits does the motion take place? (c) How fast is it moving when it is at x=7.0m? Solution: (a) Using the relationship of ( ) d d PE x Fx = − , from the diagram, the force acting on the particle at x=2.0m is the opposite of the tangent of the line L12. So 4.7N 3 17 3 ( ) d d ≈ − = − PE = x Fx (b) At the point x=2.0m, we can calculate the total mechanical energy: 2 2 4 (J) 2 1 8 2 E = PE + KE = − + × × = − As we must have E ≥ PE , PE ≤ −4J .Look at the diagram , the position limit of the particle is 1.5m ≤ x ≤ 14m . (c) 2 4 13 (m/s) 2 1 17 2 E = PE + KE ⇒ − + × × v = − ⇒ v = 3. A projectile is launched at initial speed 0 v at an angle θ with the horizontal direction. Neglect air friction and assume the trajectory is confined to a region near the surface of the Earth. Find the maximum vertical height of the projectile at the top of its flight path by using CWE theorem. Solution: Set up the coordinate system shown in figure, choose the elevation of the y –coordinate origin. The initial kinetic energy is 2 0 2 1 KE mv i = The initial potential energy is PEi = mgyi = 0 The kinetic energy at the top of its flight path is 2 0 ( cos ) 2 1 KEtop = m v θ The potential energy at the top of its flight path is PEtop = mgytop Making the appropriate substitution into the CWE theorem, y x Fig.6 0 5 10 15 -20 -15 -10 -5 0 U(x) (J) x (m) 1 2
Wother =(KEto+ PEon)-(KE+ PE) 0=[m(vo cos0)+myron]-(mvo) y 4. The total force on a system of mass m varies with position according to F= Fo (a)Calculate the work done on the system by this force as the system moves from x=0 m to x=I2 (b)Calculate the work done on the system by this force as the system moves from x =1/2 to x=0 m (c)Is the force conservative? If so, proceed with the following questions. If not, you have the rest of the afternoon off (d)Show that the following potential energy function is appropriate for this force F PE(x) Solution (a) The work done on the system by this force as the system moves from x=0 m to x=/2 is H(0→2)=F4=!Fcox2)x=0 (b)The work done on the system by this force as the system moves from x=2 to x=0 m is W(2→0)=JFd=oxm=0 (c) According to the definition of the conservative force W1+W=0 So the force is conservative dPe(x)d Fo/:2z dx2r sin(1)]= Fo cos( dPe(x) d Thus the potential energy function is appropriate for this force
( ) ( ) Wother = KEtop + PEtop − KEi + PEi We get ) 2 1 ( cos ) ] ( 2 1 0 [ 2 0 2 m v0 mgy mv = θ + top − Solving for top y θ2 2 0 sin 2g v ytop = . 4. The total force on a system of mass m varies with position according to i l x F F ˆ 2 cos 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = r π . (a) Calculate the work done on the system by this force as the system moves from x = 0 m to x = l/2. (b) Calculate the work done on the system by this force as the system moves from x = l/2 to x = 0 m. (c) Is the force conservative? If so, proceed with the following questions. If not, you have the rest of the afternoon off. (d) Show that the following potential energy function is appropriate for this force: ) 2 sin( 2 ( ) 0 l F l x PE x π π = − Solution: (a) The work done on the system by this force as the system moves from x = 0 m to x = l/2 is )d 0(J) 2 ) d cos( 2 (0 2 0 1 → = ⋅ = 0 = ∫ ∫ l x l x F r F l W v v π (b) The work done on the system by this force as the system moves from x = l/2 to x = 0 m is )d 0(J) 2 0) d cos( 2 ( 0 2 2 → = ⋅ = 0 = ∫ ∫l x l x F r F l W v v π (c) According to the definition of the conservative force: 0 QW1 +W2 = ∴ ⋅ d = 0 ∫ F r v v So the force is conservative. (d) ) 2 )] cos( 2 sin( 2 [ d ( ) d 0 0 l x F l F l x dx dx PE x π π π Q− = = dx PE x F d ( ) ∴ = − Thus the potential energy function is appropriate for this force
