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h10_6a

# h10_6a - Problem 1 a W=| F| | s| cos and =90 so W=0 b 0<90...

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Problem 1) a) W = | F | | s | cos( θ ) and θ =90 ° so W=0 b) 0 ° < θ <90 ° and so cos( θ )>0 so that W>0 c) Here θ =-180 so W<0 d) θ =0 so W>0 e) W< 0 f) θ =-90 so W=0 g) cos( θ )>0 for 0 ° > θ > -90 ° h) a. | s | =160m , | F | =18N, θ =0 so W=2900J b. | F | =30N, θ =30 so W=4200J c. | F | =12N, θ =-180 so W=-1900J d. | F | =15N, θ =-140 so W=1800J Problem 2) (7.7) perpendicular to each other. C is incorrect because work also depends on the direction of displacement Problem 3) a) Work is equal to the change in kinetic energy. Since the force on both blocks is the same and the distance is the same, an equal amount of work is done on each block. Therefore they have the same kinetic energy.

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b) Setting the kinetic energies of the two blocks equal to each other, we find that the lighter block is traveling twice as fast as the heavier one. c) Because the heavier block has four times the mass of the lighter block, when the two blocks travel with the same speed, the heavier block will have four times as much kinetic energy. The work-energy theorem implies that four times more work must be done on the heavier block than on the lighter block. Since the same force is applied to both blocks, the heavier block must be pushed through four times the distance as the lighter block. Problem 4) (7.26) a) The length the spring extends is x=-2.2 cm. F=25N. The spring formula is F=- kx so k=1140 N/m b) W=.5kx 2 =.276 J c) From the spring formula, d= 21.4 cm Problem 5) a) The work-energy theorem states that a force acting on a particle as it moves over a distance changes the kinetic energy of a particle b) To calculate the change in energy, you must know the force as a function of
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h10_6a - Problem 1 a W=| F| | s| cos and =90 so W=0 b 0<90...

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