chap8-1 - CHAPTER 8 CONSERVATION OF ENERGY ActivPhysics can...

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Unformatted text preview: CHAPTER 8 CONSERVATION OF ENERGY ActivPhysics can help with these Problems: All Activities in Section 5, Work and Energy Section 8-1: Conservative and Nonconservative Forces Problem 1. Determine the work done by the frictional force in moving a block of mass m from point 1 to point 2 over the two paths shown in Fig. 8-26. The coefficient of friction has the constant value μ over the surface. (The diagram lies in a horizontal plane.) 2 ( b ) 1 ( a ) figure 8-26 Problems 1, 2. Solution Figure 8-26 is a plane view of the horizontal surface over which the block is moved, showing the paths ( a ) and ( b ). The force of friction is μmg opposite to the displacement( f · d r = − fdr ), so W ( a ) = − μmg ( ℓ + ℓ ) = − 2 μmgℓ, and W ( b ) = − μmg √ ℓ 2 + ℓ 2 = − √ 2 μmgℓ. Since the work done depends on the path, friction is not a conservative force. Problem 2. Now take Fig. 8-26 to lie in a vertical plane, and find the work done by the gravitational force as an object moves from point 1 to point 2 over each of the paths shown. Solution Take the origin at point 1 in Fig. 8-26 with the x-axis horizontal to the right and the y-axis vertical upward. The gravitational force on an object is constant, F g = − mg ˆ , while the paths are (a) d r=ˆ dy for x = 0 and ≤ y ≤ ℓ, followed by d r=ˆ ı dx for y = ℓ and ≤ x ≤ ℓ, and (b) d r=ˆ ı dx + ˆ dy = ( ˆ ı+ˆ ) dy, for 0 ≤ y ≤ ℓ (since x = y along this path). The work done by gravity (Equation 7-11) is W ( a ) g = integraldisplay F g · d r = integraldisplay ℓ ( − mg ˆ ) · ˆ dy + integraldisplay ℓ ( − mg ˆ ) · ˆ ı dx = − mg integraldisplay ℓ dy + 0 = − mgℓ, and W ( b ) g = integraldisplay ℓ ( − mg ˆ ) · ( ˆ ı +ˆ ) dy = − mg integraldisplay ℓ dy = − mgℓ. Of course, these must be the same because gravity is a conservative force. Problem 3. The force in Fig. 8-22 a is given by F= F ˆ , where F is a constant. The force in Fig. 8-22 b is given by F= F ( x/a ) ˆ , where the origin is taken at the lower left corner of the box, a is the width of the square box, and the distance x increases horizontally to the right. Determine the work done by F on an object moved counterclockwise around each box, starting at the lower left corner. ( a ) ( b ) figure 8-22 Problem 3. Solution The path around the square consists of four segments, each along a side, in the direction of a counterclock- CHAPTER 8 99 wise circulation. Thus W = contintegraldisplay F · d r= integraldisplay a F ( y = 0) · ˆ ı dx + integraldisplay a F ( x = a ) · ˆ dy + integraldisplay a F ( y = a ) · ˆ ı dx + integraldisplay a F ( x = 0) · ˆ dy (Note that a path parallel to the x-axis from right to left is represented by d r=ˆ ı dx with x going from a to 0 in the limits of integration, etc.) (a) For F= F ˆ , the expression for the work becomes W = 0 + F integraldisplay a dy + 0 + F integraldisplay...
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This note was uploaded on 04/24/2010 for the course MATH 9374 taught by Professor Smith during the Spring '10 term at Berkeley.

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chap8-1 - CHAPTER 8 CONSERVATION OF ENERGY ActivPhysics can...

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