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A mass of 0.368 kilograms is suspended from a spring, and the spring stretches 12.3 centimeters. What work must be done to stretch the spring twice as far as it is stretched in part b? Include units in your answer. Answer must be in 3 significant digits.

Sagot :

The gravitational force acting on the mass is,

[tex]F=mg[/tex]

The spring constant of the spring can be given as,

[tex]k=\frac{F}{d}[/tex]

Substitute the known expression,

[tex]k=\frac{mg}{d}[/tex]

Substitute the known values,

[tex]\begin{gathered} k=\frac{(0.368kg)(9.8m/s^2)}{(12.3\text{ cm)(}\frac{1\text{ m}}{100\text{ cm}})}(\frac{1\text{ N}}{1kgm/s^2}) \\ \approx29.3\text{ N/m} \end{gathered}[/tex]

The work done to stretch the spring is,

[tex]W=\frac{1}{2}k(2x)^2[/tex]

Here, x is the amount of spring stretched in part (b).

Substitute the known values,

[tex]\begin{gathered} W=\frac{1}{2}(29.3N/m)(2(0.165m))^2(\frac{1\text{ J}}{1\text{ Nm}}) \\ \approx1.60\text{ J} \end{gathered}[/tex]

Thus, the work done to stretch the spring is 1.60 J.

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