chap 6 - WORK, ENERGY, AND POWER 6 EXERCISES Section 6.1...

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6.1 WORK, ENERGY, AND POWER EXERCISES Section 6.1 Work 12. If the force is constant and parallel to the displacement, WFrF r == = = v v ⋅∆ () . 75 12 900 Nm J 13. INTERPRET The relevant physical quantity here is the work done by a person sliding a box. DEVELOP If you push parallel to a level floor, the applied force equals the frictional force (since the acceleration is zero), Ff a = k . Since the normal force equals the weight of the box: nm g = , the applied force is Fn m g a µµ kk Equation 6.1, WF x =∆ , can then be used to find the work done. EVALUATE The applied force is constant and parallel to the displacement. So using Equation 6.1, we get W F r f rn rm g r aa === = = ∆∆ k kg ( . )( )( . 021 50 98ms m J /) ( . ) 2 4 8 494 = ASSESS The units are correct, 11 1 2 JN mk g m s m . =⋅ = / If the floor were frictionless , µ k = 0 then the work done would be W a = 0, as expected. 14. Lifting the beam at constant speed, the crane exerts a constant force vertically upward and equal in magnitude to the weight of the beam. During the horizontal swing, the force is the same, but is perpendicular to the displacement. The work done is v v Fr m g j y j x i m g y + ( ˆ ˆˆ ) ( . kg 650 9 8 23 147 2 ms m kJ ( ) . = 15. INTERPRET This problem is about the work done by gravity on the water falling from a certain height. DEVELOP Since the density of water is 1000 kg/m 3 , the mass of a cubic meter of water is 1000 kg, and the force of gravity at the Earth’s surface on a cubic meter of water is Fm g g = ( . ) 1000 9 8 9800 kg m/s N 2 vertically downward. We can then use Equation 6.1, x , to find the work done. EVALUATE The displacement of the water is parallel to this, so the work done by gravity on the water is y gg = × ( ) . 9800 980 9 6 10 6 J ASSESS The units are correct, m . The greater the distance the water falls, the larger the amount of work done by gravity. 16. The average force exerted by the meteorite parallel to its penetration into the ground is FW x av MJ/ / 140 0 75 187 ., mM N = or about 21,000 tons. 17. INTERPRET This problem is about the work done by the elevator cable on the elevator as it accelerates upward. DEVELOP To give the elevator a constant upward acceleration a y , the tension in the cable must satisfy Tm gm a ga yy −= = + We can then use Equation 6.1, x , to find the work done. EVALUATE Acting over a parallel displacement yh = upward, the amount of work done by the cable on the elevator is W Tym ga ym g g h m g h Ty ==+ =+ = ( . ) . 01 6
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6.2 Chapter 6 ASSESS The units are correct, 11 JN m . =⋅ The greater the upward acceleration a y , the larger the amount of work must be done by the cable. Of course, if the elevator undergoes free fall with ag y =− , then the tension in the cable is zero, and no work is done on the elevator. 18. This follows easily from the definition of the scalar product in terms of components: v v v ABC AB C AB C A xx x yy y zz z x ⋅+ () ( ) ( ) ( ) =+ ++ = B A BA CA C xy yz zx z +++++= v v vv .
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chap 6 - WORK, ENERGY, AND POWER 6 EXERCISES Section 6.1...

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