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, amFamF ′=′== 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 )()( lxkxxF x ?= for lx ≤≤0 , 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) 0=d . (B) 4/ld = . (C) 2/ld = . (D) 2/ld = . (E) ld = Solution: The force is 4 ) 2 ()()( 2 2 kll xklxkxxF 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 mvKE = , 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: KEW total ?= , the net work done on the puck is zero. 5. Which of the following forces is not conservative? （ C ） (A) jiF ? 4 ? 3 += r (B) jyixF ? 4 ? 3 += r (C) jxiyF ? 4 ? 3 += r (D) jyixF ? 4 ? 3 22 += r Solution The work done by the force in (C) is ∫∫∫ +=? f i f i x x y y yxxyrF d4d3d 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 PEW ??= conserv , if we choose the zero potential energy on the table the work required to pull the hanging part back on the table is mgLLmgmghW 32 1 8 1 4 1 conserv =?== 2 A force on a system has only an x-component F x ; 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 m0= i x and m10= f x J5.2355.23575)m5)(N30( 2 1 area =?=== Wππ 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) F x (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 1.11032.98255.2 2 1 )6.43(0 2 2 =?=??? ++?= xxx kxxmg 4. The work done by the gravitational force in moving a system of mass m at a distance r i from the center of a uniform sphere (of mass M) to a distance r f is ) 22 ( 2 2 3 i f r r R GMm ?? , where both r i and r f 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 rr R M r Gm F G ?? 3 4 3 4 3 3 3 2 ??=???= π π r ) 22 ( 2 dr?)(d 2 2 3 2 33 | i f r r r r f i G r r R GMmr R GMm rrr R GMm rFW f i f i ??=??=???=?=? ∫∫ rr r 5. A particle moves along the x axis under the influence of a conservative force that is described by ixeF x ? 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 2222 22 d 2 dd)( 2 x x x x x x x p eexexxerFxU ββββ β α β αα α ? ∞ ? ∞ ? ∞ ? ∞ =?=?=?=?= ∫∫∫ v r 6. A force jkyikxF ?? ??= 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 22 yxk + . m R Fig.4 i r r f r r Solution: (a) Assume that the particle moves from the position ) ? ?? ( kzjyixr iiii ++ r to the position ) ? ?? ( kzjyixr ffff ++ r . Thus the work done by the force jkyikxF ?? ??= r is ∫∫∫ ????=?+?=?= f i f i f i x x y y ifif r r yykxxkjkyikxrFW )( 2 1 )( 2 1 j ? dy) ? (i ? dx) ? (d 2222 r 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 [ 2222 ifiiff PEPEyxkyxkW ??=+?+?= Potential energy function for this force is )( 2 1 22 yxkPE += 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 ? ? ? =? =? 0cos 0sin θ θ Tmg TF Since θtanmgF = 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 )J(24.3548)182.0(2581.980)35cos( =?×××?=??=?= llmgEW F o F r 35° y x 0 T v gm v 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 F x ?= , from the diagram, the force acting on the particle at x=2.0m is the opposite of the tangent of the line L 12 . So N7.4 3 317 )( d d ≈ ? =?= PE x F x (b) At the point x=2.0m, we can calculate the total mechanical energy: )J(422 2 1 8 2 ?=××+?=+= KEPEE As we must have PEE ≥ , J4?≤PE .Look at the diagram , the position limit of the particle is m14m5.1 ≤≤ x . (c) )m/s(1342 2 1 17 2 =??=××+??+= vvKEPEE 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 mvKE i = The initial potential energy is 0== ii mgyPE The kinetic energy at the top of its flight path is 2 0 )cos( 2 1 θvmKE top = The potential energy at the top of its flight path is toptop mgyPE = Making the appropriate substitution into the CWE theorem, y x Fig.6 051015 -20 -15 -10 -5 0 U (x) (J) x (m) 1 2 )()( iitoptopother PEKEPEKEW +?+= We get ) 2 1 (])cos( 2 1 [0 2 0 2 0 mvmgyvm top ?+= θ Solving for top y θ 2 2 0 sin 2g v y top = . 4. The total force on a system of mass m varies with position according to i l x FF ? 2 cos 0 ? ? ? ? ? ? = π r . (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 xlF xPE 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 )J(0d) 2 cos(d) 2 0( 2 0 01 ==?=→ ∫∫ l x l x FrF l W πv v (b) The work done on the system by this force as the system moves from x = l/2 to x = 0 m is )J(0d) 2 cos(d)0 2 ( 0 2 02 ==?=→ ∫∫ l x l x FrF l W πv v (c) According to the definition of the conservative force: 0 21 =+WWQ 0d =?∴ ∫ rF v v So the force is conservative. (d) ) 2 cos()] 2 sin( 2 [ d)(d 0 0 l x F l xlF dxdx xPE ππ π ==?Q dx xPE F )(d ?=∴ Thus the potential energy function is appropriate for this force.