T1T222111222221122222112122222EEmvmgymvmgyvgyvgyvvgygy=+=++=+=+−Sincey2= 0, 2gy2= 0, therefore:22112222(9.7 m/s)2(9.80 m/s)(4.2 m)13 m/svvgyv=+=+=The skier’s speed upon touching the hillside is 13 m/s.5.∆y= 4.4¯102mv2= 93 m/sv1= ?T1T222112222112222122121221221112222222 ()(93 m/s)2(9.8 m/s)(0440 m)5.0 m/sEEmvmgymvmgyvgyvgyvvgygyvvg yyv=+=++=+=+−=+−=+−=The speed of the water at the top of the waterfall is 5.0 m/s.6.v1= 9.7 m/s∆y= 4.7 mv2= ?

Copyright © 2003 NelsonChapter 4Work and Energy251T1T222112222112222211222112222112222222 ()(9.7 m/s)2(9.8 m/s)(04.7 m)1.4 m/sEEmvmgymvmgyvgyvgyvvgygyvvg yyv=+=++=+=+−=+−=+−=The cyclist crests the hill at a speed of 1.4 m/s.7.v2= ?First determine how high the pendulum is vertically raised:2221221124.5 cm85.5 cm85.5 cm24.5 cm81.915 cmhhh+==−=1221285.5 cm85.5 cm85.5 cm81.915 cm3.585 cmhhhhh+==−=−=Using the valueh2= 3.585 cm, or 0.03585 m, calculate the maximum speed:T1T2221122121222212212221122since022222 ()2(9.80 m/s)(0.03585 m0)0.838 m/sEEmvmgymvmgyvgyvgyvgygyvg yyv=+=+==+=−=−=−=The maximum speed of the pendulum bob is 0.838 m/s.

Copyright © 2003 NelsonChapter 4Work and Energy252Applying Inquiry Skills8.∆y(m)∆Eg(J)EK(J)ET(J)8.003920.003926.0029498.03924.001961963922.0098.02943920.000.00392392Making Connections9.A wrecking ball works as a pendulum that is slowly pulled back, increasing its gravitational energy. When it is released,the gravitational potential energy is converted into kinetic energy which is used to destroy buildings.PRACTICE(Page 200)Understanding Concepts10. (a)The energy supplied becomes sound and thermal energy through friction.(b)The energy supplied still produces sound and thermal energy, but some is also converted into kinetic energy.11.FK= 67 N∆d= 3.5 m(a)W= ?2(cos)(67 N)(cos180)(3.5 m)2.310JWFdWθ=∆=°= −×The amount of work done by friction is –2.3¯102J.(b)Eth= ?thK2th(67 N)(3.5 m)2.310JEFdE=∆==×The amount of thermal energy produced is 2.3¯102J.12.Eth= 0.620 JFK= 0.83 N∆d= ?

Copyright © 2003 NelsonChapter 4Work and Energy253The 0.620 J of energy comes from the work done by friction, therefore:thKthK0.620 J0.83 N0.75 mEFdEdFd=∆∆==∆=The plate slides 0.75 m.13.m= 22.0 kgF= 98 NFK= 87 Nvi= 0.0 m/s∆d= 1.2 mvf= ?thk2K22K2K221(cos)2(cos)12(cos)0.5(98 N)(cos0)(1.2 m)87 N(1.2 m)0.5(22.0 kg)1.1 m/sWEEFdFdmvFdFdvmFdFdvmvθθθ=+∆=∆+∆−∆=∆−∆=°−==The speed of the cabinet after it moves 1.2 m is 1.1 m/s.14.m= 0.057 kg∆d= 25 cm = 0.25 mvf= 5.7 cm/s = 5.7¯10–2m/sFK= 0.15 Nvi= ?K1thK2221K22K2212K2122111221212120.51(0.15 N)(0.25 m)(0.057 kg)(5.710m/s)20.5(0.057 kg)1.1 m/sEEEmvFdmvFdmvvmFdmvvmv−=+=∆+∆+=∆+=+×==The initial speed of the pen is 1.1 m/s.Applying Inquiry Skills15. (a)The law of conservation of energy could be verified by observing how close to its original position the pendulumwould return after a swing.

Copyright © 2003 NelsonChapter 4Work and Energy254(b)Some sources of error would be loss of energy due to friction at the point of attachment and air friction of the movingpendulum. Some error may be observed if the string is somewhat elastic.

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