L06_hw - Chapter 6 Homework 5 Ed. Q: 1, 2, 5, P:1, 3, x,...

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Unformatted text preview: Chapter 6 Homework 5 Ed. Q: 1, 2, 5, P:1, 3, x, 13*, 19, 28* 6 Ed. Q: 1, 2, 5, P: 1, 3, 7, 13*, 17, 25* 6 Ed. Q: 9, 13, 21 P:29, 37, 40, 42, 44, 47, 58 P1. How much work is done by the gravitational force when a 265 kg pile driver falls 2.80 m? Solution: The force points "down". The direction of the displacement is "down". The angle between the two is zero cos kg m/s m cos J P3. A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a) 4.0 m along the floor against a friction force of 230 N (b) 4.0 m vertically? Solution: (a) (b) d F F Ff r d mg (a) Force is along the floor. Displacement along the floor Work done by the pulling force: N m cos J (b) Force is "up". Displacement is "up" Newton's Second Law along the vertical: N N angle is 0 angle is 0 Work done by the pulling force: N m cos 1 P7. A lever such as that shown in Fig.6-35 can be used to lift objects we might not otherwise be able to lift. Show that the ratio of output force, F to input force, F is related to the lengths and from the pivot point by (ignoring friction and the mass of the lever), given that the work output equals work input. Solution: li hi ho lo The two triangles are similar Work done by the input force: Work done by the output force: Work in = Work out P17. How much work is required to stop an electron ( moving with a speed of 1.90 m/s. Solution: Initial kinetic energy of the electron is: kg m/s kg which is J Final kinetic energy of the electron is: !" #$ % " ! "%& ) J '" ( J ! " !% " ! "% ' "! " $ J by a P25*. A 285-kg load is lifted 22.0 m vertically with an acceleration single cable. Determine (a) the tension in the cable (b) the net work done on the load (c) the work done by the cable on the load (d) the work done by gravity on the load (e) the final speed of the load assuming it started from rest. Solution: 2 d T mg (a) Tension in the cable ? Newton's Second Law: * + + * !"% * + , m/s , m/s , N kg (b) Work done by the net force? Net force is "up", displacement is "up", angle is 0 N ! "% cos (c) Work done by cable? Tension points "up", displacement is "up", angle is 0 + cos N m 7.13 ' #" (d) Work done by gravity on the load? Gravity "down", displacement "up" angle is 180 cos kg m/s cos (e) Final speed of the load ( ! " ! m/s J !" #$ - + '" m/s P29. A 1200-kg car rolling on a horizontal surface has speed km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring? 3 Solution: km/h . "& 2 2 " % "' J , m/s / ! 0 & %" % " ' 3 2 2 3 /% 1/ J , kg m/s m N m P37. A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m/s (a) How fast is he going as he lands on the trampoline, 3.0 m below? (b) If the trampoline behaves like a spring with spring stiffness constant 6.2 N/m, how far does he depress it? Solution: Choose to measure the height from the level of the trampoline Initial height is y 3.0 m At the top of the platform 2 $ %$ 2 /1 ! Landing on the trampoline 2 %$ 2 /1 ! At the bottom of the trampoline 2 $ %$ 2 /1 ! $ J , $ , J , 2 2 2 J , $ , $ Only gravity and the elastic force are acting, and they both are conservative conserved! (a) velocity at landing ? !" $ %% " % 1 !" $ 4 "! ! ! ! % " % $ , $ , m/s m , m/s (b) how far the trampoline compresses ? $ !" $ %% " % 1 !" $ %% " # %% ? Energy is 1 m/s !" 4 $ , $ , $ kg 6.2 m/s N/m $ , $ 5 $ , $ $ , 5 $ $ 6 $ must be negative, that is, it must be below the level of the trampoline, thus $ m P40. A block of mass slides without friction along the looped track shown on the diagram. If the block is to remain on the track, even at the top of the circle (whose radius is r), from what minimum height h must it be released? Solution: a FN mg h kg 6.2 m/s N/m kg m/s 6.2 m, N/m kg m/s If the block is to remain on the track, the centripetal acceleration must remain large enough, so that the net force is always equal or larger than that of gravity. At the top of the track, two forces act on the block: Newton's Second Law: , 7 - '1 - 7 (*) 7 The velocity at the top depends on the initial height. We are going to apply the conservation of energy principle: ( ! " ! !" $ ! " #$ ! ! ' ! /" % "& '"/ The only two forces acting on the block are gravity and the normal force (between the track and the block). 1. Gravity is conservative force. 5 2. The normal force never does any work, because it is always perpendicular to the direction of motion. Therefore, the work done by the non-conservative forces is zero and the energy is conserved. ( 2 2 J , ! " ! !" $ ! " $ %% " ! % " % 2 ! " $ %% " % 1 & % " % ' 2 , , !" $ %% " ! % , Substituting in (*) 7 7 7 ! " % ! " $ %% " % 1 & % " % ' P42. A 62-kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12m, and falls a total of 31 m (a) Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies (b) Calculate the maximum acceleration she experiences Solution: Hooke's Law: /% "%' ! & % " ' )3 ) 3, where ) 3 is the stretching of the cord % % "! % 8!/% "%' " "! % (a) Two forces act on the jumper - gravity and tension from the cord. If we can assume Hooke's law is valid and treat the cord as a spring - all forces acting on the jumper will be conservative Mechanical energy is conserved. Let us count the vertical position from the unstretched position of the hanging rope, that is 12 m beneath the bridge. So, we have: $ m $ m 6 Energy at the top of the jump 2 $ 2 " /% ' Energy at the bottom of the jump 2 $ 2 " /% ' )3 J J , 2 , J 2 , 2 $ , $ , m/s )3 )3 2 $ $ kg kg , m m, m m kg m/s N/m m , m m/s (b) At the top of the bridge, the jumper falls freely with an acceleration of m/s . As the rope begins to stretch, the jumper begins to experience an elastic force (tension) pulling her upwards. The elastic force changes and it is at its maximum at the bottom of the jump. Newton's Second Law: "up" is "+" The "-" in the Hooke's law determines the direction in which the force acts. Since we have already taken this into consideration (the elastic force is up in the Newton's Second Law equation), we write the force simply is ) 3 in the following equation: )3 N/m m kg m/s m/s kg P44. A 0.40-kg ball is thrown with a speed of 12 m/s at an angle of 33. (a) What is its speed at is highest point, and (b) how high does it go? Ignore air resistance. Solution: (a) At the highest point, the velocity of the ball has only a horizontal component: cos m/s 3 (b) The only force acting on the ball is gravity, which is a conservative force energy is conserved. Choose the ground as zero for the potential energy. Mechanical " 7 Energy at the moment of throwing 2 J 2 , Energy at the highest point 2 3 2 , , 3 , , m/s m 3 3 m/s , m/s P47. Two railroad cars, each of mass 7650 kg and traveling 95 km/h in opposite directions, collide head-on and come to rest. How much thermal energy is produced in this collision? Solution: We do not count thermal energy in our energy balance. Thus, if any thermal energy is produced, it will appear as a loss in our mechanical energy. Before collision After collision , J 2 , 2 , J + " !" $ 9 ) 9 J P58. How long will it take a 1750-W motor to lift a 315-kg piano to a sixth-story window 16.0 m above? Solution: !" $ 2 % " %% "/ By lifting the piano, its potential energy increases. Assume constant speed throughout, so the kinetic energy is the same. Since the piano has increased its mechanical energy by , that energy must be supplied by someone - it comes from the 1750-W motor. kg m/s m !" $ % " s 2 4" 2 6 W 8 ...
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This document was uploaded on 09/21/2009.

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