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Work of 2 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (19 cm)? Don't forget to enter the correct units.
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\[Work = Force\cdot distance\] Work = 2 Joules distance = .19 meters
Work done , We = 1/2*kx^2 2joules = 1/2*k*(0.19)^2 find k , spring constant Force that holds the spring stretched = k*0.19 N
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