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CHAPTER 8: Conservation of Energy Solutions to Problems 22. (a) Draw a free-body diagram for each block. Write FN FT Newton’s second law for each block. Notice that the FT yB acceleration of block A in the yA is 0 zero. yA xA mB g Fy1 FN mA g cos 0 FN mA g cos mAg F x1 FT mA g sin mAaxA F y2 mB g FT mBa yB FT mB g a yB Since the blocks are connected by the cord, a yB a xA a. Substitute the expression for the tension force from the last equation into the x direction equation for block 1, and solve for the acceleration. mB g a mA g sin mA a mB g mA g sin mA a mBa m 9.80 m s mA sin 5.0 kg 4.0 kg sin 32 ag 3.1m s 2 B 2 mA m B 9.0 kg (b) Find the final speed of m B (which is also the final speed of mA ) using constant acceleration relationships. v 2 v0 2ay v 2 2 g 2 m B mA sin h f f mA mB m 0.75 m mA sin 5.0 kg 4.0 kg sin 32 v f 2 gh B mA mB 2 9.80 m s 2 9.0 kg 2.2 m s (c) Since there are no dissipative forces in the problem, the mechanical energy of the system is conserved. Subscript 1 represents the blocks at the release point, and subscript 2 represents the blocks when m B reaches the floor. The ground is the zero location for gravitational potential energy for mB , and the starting location for mA is its zero location for gravitational potential energy. Since m B falls a distance h, mA moves a distance h along the plane, and so rises a distance h sin . The starting speed is 0. E1 E2 0 mA gh 1 2 mA mB v 2 2 mB gh sin mB mA sin v2 2 gh mA mB This is the same expression found in part (b), and so gives the same numeric result. 36. (a) Use conservation of energy to equate the potential energy at the top of the circular track to the kinetic energy at the bottom of the circular track. Take the bottom of the track to the be 0 level for gravitational potential energy. Etop Ebottom mgr 1 mvbottom 2 2 vbottom 2 gr 2 9.80 m s2 2.0 m 6.261m s 6.3m s (b) The thermal energy produced is the opposite of the work done by the friction force. In this situation, the force of friction is the weight of the object times the coefficient of kinetic friction. E thermal Wfriction Ffriction x Ffriction x cos k mg x cos180 k mg x 0.251.0 kg 9.80 m s 2 3.0m 7.35J 7.4 J (c) The work done by friction is the change in kinetic energy of the block as it moves from point B to point C. Wfriction K KC K B 1 m vC vB 2 2 2 2Wfriction 2 7.35J vC vB 6.261m s 4.9498 m s 4.9 m s 2 2 m 1.0 kg (d) Use conservation of energy to equate the kinetic energy when the block just contacts the spring with the potential energy when the spring is fully compressed and the block has no speed. There is no friction on the block while compressing the spring. Einitial Efinal 1 2 mvcontact 1 kxmax 2 2 2 2 vcontact 4.9498 m s 2 k m 1.0 kg 612.5 N m 610 N m 2 xmax 0.20 m 2 74. First, consider a free-body diagram for the cyclist going down hill. Write Ffr FN Newton’s second law for the x direction, with an acceleration of 0 since the y cyclist has a constant speed. x F x mg sin Ffr 0 Ffr mg sin mg y Now consider the diagram for the cyclist going up the hill. Again, write x Newton’s second law for the x direction, with an acceleration of 0. FN FP F Ffr x Ffr FP mg sin 0 FP Ffr mg sin Assume that the friction force is the same when the speed is the same, so the mg friction force when going uphill is the same magnitude as when going downhill. FP Ffr mg sin 2mg sin The power output due to this force is given by Eq. 8-21, with the force and velocity parallel. P FPv 2mgv sin 2 75 kg 9.80 m s2 4.0 m s sin 6.0 o 610 W 90. (a) Draw a free-body diagram for the block at the top of the curve. Since the FN mg block is moving in a circle, the net force is centripetal. Write Newton’s second law for the block, with down as positive. If the block is to be on the verge of falling off the track, then FN 0. F R FN mg mv 2 r mg mvtop r vtop gr 2 Now use conservation of energy for the block. Since the track is frictionless, there are no non-conservative forces, and mechanical energy will be conserved. Subscript 1 represents the block at the release point, and subscript 2 represents the block at the top of the loop. The ground is the zero location for potential energy y 0 . We have v1 0, y1 h, v2 gr , and y2 2 r. Solve for h. E1 E2 1 2 mv12 mgy1 1 mv2 mgy2 0 mgh 1 mgr 2mgr 2 2 2 h 2.5 r (b) See the free-body diagram for the block at the bottom of the loop. The net force is again centripetal, and must be upwards. FN F R FN mg mv 2 r FN mg mvbottom r 2 mg The speed at the bottom of the loop can be found from energy conservation, similar to what was done in part (a) above, by equating the energy at the release point (subscript 1) and the bottom of the loop (subscript 2). We now have v1 0, y1 2h 5r , and y2 0. Solve for v 2 . E1 E2 1 2 mv12 mgy1 1 mv2 mgy 2 0 5mgr 1 mvbottom 0 2 2 2 2 vbottom 10 gr FN mg m vbottom r mg 10mg 11mg 2 2 (c) Again we use the free body diagram for the top of the loop, but now the normal force does not vanish. We again use energy conservation, with v1 0, y1 3r , and y2 0. Solve for v 2 . F R FN mg m v 2 r FN m vtop r mg 2 E1 E2 1 2 mv12 mgy1 1 mv2 mgy2 0 3mgr 1 mvtop 0 2 2 2 2 vtop 6gr FN mvtop r mg 6mg mg 5mg 2 2 (d) On the flat section, there is no centripetal force, and FN mg .
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