Answer: The work W required to stretch the spring an additional 6 centimeters
Explanation:
The force require to stretch the spring by 6 cm = 1200 dynes = \(1200\times 10^{-5} N=0.012 N\)
Displacement of from the initial position = 6 cm = 0.06 m
\(F=k\times x\)
\(0.012 N=k\times 0.06 m\)
k = 0.2 N/m
Work done is guiven by: \(\frac{1}{2}kx^2\)
Work done when spring expanded to0.0 6 cm= \(\frac{1}{2}kx^2=\frac{1}{2}\times 0.2 N/m\time (0.06)^2\)..(1)
Work done when spring again expanded to 0.06 cm that is 0.12 m (0.06+0.06 m) = \(\frac{1}{2}kx^2=\frac{1}{2}\times 0.2 N/m\time (0.12)^2\)..(2)
Work done(W) required to stretch the spring an additional 0.06 m: (2)-(1)
\(W=\frac{1}{2}0.2 N/m\times (0.12 m)^2-\frac{1}{2}0.2 N/m\times(0.06 m)^2=0.00108 Joules\)
The work W required to stretch the spring an additional 6 centimeters
