Ch_7Word - Chapter 7 CHAPTER 7 Work and Energy 1 2 3 The...

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Chapter  7 CHAPTER 7 - Work and Energy 1. The displacement  is in the direction of the gravitational force, thus W  =  Fh   cos 0 °  =  mgh  = (250 kg)(9.80 m/s 2 )(2.80 m) =       6.86 × 10 3  J . 2. The displacement  is opposite to the direction of the retarding  force, thus W  =  Fx  cos 180 °  = (535 N)(1.25 × 10 3  m)(– 1) =       – 6.69 × 10 5  J . 3. Because there is no acceleration, the contact force must have the same magnitude  as the weight.   The displacement  in the direction of this force is the vertical displacement.  Thus,  W  =  F  ? y  = ( mg ) ? y  = (65.0 kg)(9.80 m/s 2 )(20.0 m) =       1.27 × 10 4  J . 4. ( a ) Because there is no acceleration, the horizontal applied  force must have the same magnitude  as  the friction force.  Thus,  W  =  F   ? x  = (230 N)(4.0 m) =       9.2 × 10 2  J . ( b ) Because there is no acceleration, the vertical applied  force must have the same magnitude  as  the weight.  Thus,  W  =  F   y  =  mg   y  = (1200 N)(4.0 m) =        4.8 × 10 3  J . 5. Because there is no acceleration, from the force diagram  we  see that F N  =  mg ,   and     F  =  F fr  =  μ k mg . Thus,  W   =  F   x  cos 0 °  =  μ k mg   x  cos 0 °   = (0.50)(160 kg)(9.80 m/s 2 )(10.3 m)(1) =       8.1 × 10 3  J . 6. Because the speed  is zero before the throw  and  when  the rock reaches the highest point, the positive  work of the throw  and  the (negative) work done by the (downward)  weight must add  to zero.  Thus,  W net  =  W throw  +  mgh  cos 180 °  = 0,   or    h  = –  W throw / mg  cos 180 °  = – (80.0 J)/(1.85 kg)(9.80 m/s 2 )(–1) =       4.41 m . 7. 1 J = (1 kg     m/s 2 )(1 m)(1000 g/kg)(100 cm/m) 2  = 1 × 10 7  g     cm/s 2  =        1 × 10 7  erg . 1 J = (1 N     m)(0.225 lb/N)(3.28 ft/m)  =         0.738 ft     lb . 8. The maximum  amount  of work that the hammer  can do is the work that was done by the weight as the  hammer  fell: W max  =  mgh  cos 0 °  = (2.0 kg)(9.80 m/s 2 )(0.50 m)(1) =       9.8 J . People add  their own force to the hammer  as it falls in order that additional work is done before the  hammer  hits the nail, and  thus more work can be done on the nail. 9. If we assume  the width  of a cut is 0.30 m, and  we cut the lawn  parallel to the long side, the number  of  cuts required  is N  = (10.0 m)/(0.30m/cut)  = 33 cuts, each 20 m long. Thus the total work (ignoring turn-arounds) is W  =  Fx  = (15 N)(33)(20 m) =       1.0 × 10 4  J .
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