chap 7 - CONSERVATION OF ENERGY 7 EXERCISES Section 7.1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
7.1 CONSERVATION OF ENERGY EXERCISES Section 7.1 Conservative and Nonconservative Forces 11. INTERPRET In this problem we want to find the work done by the frictional force in moving a block from one point to another over two different paths. DEVELOP Figure 7.16 is a plan view of the horizontal surface over which the block is moved, showing the paths (a) and (b) . The force of friction is fnm g == µµ opposite to the displacement. Since f is constant, using Equation 6.1, the work done is Wfr f r =⋅ = ± ± ∆∆ EVALUATE The work done by friction along path (a) with rLL L a =+= 2 is Wf r m g L m g L aa =− () 22 Similarly, the work done by friction along path (b) with rL L L b =+ = 2 is r m g L m g L bb ASSESS Since the work done depends on the path chosen, friction is not a conservative force. 12. Take the origin at point 1 in Fig. 7.16 with the x axis horizontal to the right and the y axis vertical upward. The gravitational force on an object is constant, ² Fm g j g ˆ , while the paths are (a) dr jdy ² = ˆ for x = 0 and 0 ≤≤ yL , followed by idx ² = ˆ for = and 0 xL , and (b) idx jdy i j dy ² ˆˆ ( ), ++ for 0 (since xy = along this path). The work done by gravity (Equation 6.11) is W F mg j mgj ndx g a g L ( ˆ ) ˆ ( ˆ ) =⋅ = + ∫∫ ² ²² ⋅⋅ 000 0 LL mg dy mgL + =− and W mg j n j dy mg dy mgL g b L ( ˆ )( ˆ ) + = = ² 00 ³ Of course, these must be the same because gravity is a conservative force. Section 7.2 Potential Energy 13. INTERPRET The problem is about gravitational potential energy relative to a reference point of zero energy. In Example 7.1, the reference point was taken to be the 33rd floor. In this problem, we take the street level to be our reference point. DEVELOP The change in potential energy with a change in the vertical distance y is given by Equation 7.3, Um g y = . Each floor is 3.5 m high. EVALUATE (a) The office of the engineer is on the 33rd floor, or is 32 stories above the street level (the first floor) where U 1 0 = .Thus, the difference in gravitational potential energy is UU U U m gy = × = 33 1 33 05 5 9 8 3 2 3 5 Nm (. ) ) 6 0 4 .k J (b) At the 59th floor, UU U × = 59 1 55 9 8 58 3 5 109 ) ) . k J (c) Street level is the zero of potential energy, U 1 0 = . 7
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
7.2 Chapter 7 ASSESS Potential energy depends on the reference point chosen, but potential energy difference between two points does not. What matters physically is the difference in potential energy. The differences in potential energy between any two levels are the same as in Example 7.1, e.g., UU 59 33 109 60 4 −= = (. ) kJ kJ 49.0 kJ. 14. If we define the zero of potential energy to be at zero altitude () , y = 0 then U , 00 = and Equation 7.3 (for the gravitational potential energy near the surface of the Earth, ||< < km y 6370 ) gives Uy U mgy ( ) −== = mgy . Therefore, (a) U ( . ) . , 1900 70 9 8 1900 1 30 mN m M J = and (b) U ) × 86 70 9 8 86 m 59 0 .. kJ 15. INTERPRET The problem is about the change in gravitational potential energy as a person comes down from the mountaintop.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at Berkeley.

Page1 / 18

chap 7 - CONSERVATION OF ENERGY 7 EXERCISES Section 7.1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online