Question

When a certain spring is expanded 6 centimeters from its natural position and held fixed, the force necessary to hold it is 1200 dynes. Find the work W required to stretch the spring an additional 6 centimeters

Answer 1

The force on a spring isF= k (d1 - d0)we say d0=0 and d1 is 6 cm so 1200= k * 6200= k
F= k ( d1- d0) where d1 is 12 cm and d0 is 6 cmF to stretch additional 6 cm is 200* 6F= 1200where work is F* change in distancewe changed the distance by 6 cm so it should be 1200 * 6= 7200

Answer 2

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