Yefim S. answered • 05/19/23
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W = FΔx/2; F = 2W/Δx = 2·5 J/0.2m = 50 N
Lily K.
asked • 05/19/23Spring stretching and joules
Work of 5 Joules done in stretching a spring from its natural length to 20 cm beyond its natural length.
What is the force (in Newtons) that holds the spring stretched at the same distance (20 cm)? (include units)
Answer=?
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NORMAN E.
Why did you divide by 2? There's no averaging here. Work dne equals (constant) force times distance. The answer is 25 newtons.05/24/23