If you stand on a bathroom scale, the spring inside the scale compresses 0.60 mm, and it tells you your weight is 760 N. Now if you jump on the scale from a height of 1.0 m, what does the scale read at its peak?

When a spring of spring constant k is compressed or stretched by a distance x, restoring force develops in the spring, which tries to move the spring back to its original position. The direction of the restoring force is opposite to the displacement, and its magnitude is given as:

\(F = kx\)

The spring is stretched from its equilibrium length by a distance:

\(\begin{aligned}x &= 0.66\;{\rm{mm}}\\ &= 0.60\;{\rm{mm}} \times \left( {\frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right)\\ &= 0.60 \times {10^{ - 3}}\;{\rm{m}}\end{aligned}\)

Your weight on the scale is \(W = 760\;{\rm{N}}\).

The height of your jump above the scale is h = 1.0 m.

If k is the spring constant of the spring, the magnitude of the restoring force is:

\(F = kx\)

When you stand on a bathroom scale, your weight acts downward and the spring gets compressed, due to which it exerts a restoring force in an upward direction, thereby showing your weight on the scale. Thus, the magnitude of the restoring force must be equal to your weight, i.e.,

\(\begin{aligned}W &= F\\W &= kx\\k &= \frac{W}{x}\\ &= \frac{{760\;{\rm{N}}}}{{0.60 \times {{10}^{ - 3}}\;{\rm{m}}}}\\ &= 12.67 \times {10^5}\;{\rm{N/m}}\end{aligned}\)

Thus, the spring constant of the spring is \(12.67 \times {10^5}\;{\rm{N/m}}\).

Suppose the equilibrium position of the spring is taken as the reference position and the downward direction as positive. In that case, thecompression in the spring will be positive, and the height above the spring will be negative.

Let the maximum compression in the spring be \(x'\) when you jump on the bathroom scale from height h. Thus, the elastic potential energy of the springis:

\(P{E_{{\rm{el}}}} = \frac{1}{2}k{x'^2}\)

The gravitational potential energy is\(P{E_{\rm{G}}} = - m{\rm{g}}h\).

Since the total energy of the system at any point should remain constant, therefore:

\(\begin{aligned}\frac{1}{2}k{{x'}^2} &= m{\rm{g}}h\\\frac{1}{2}k{{x'}^2} &= Wh\\{{x'}^2} &= \frac{{2Wh}}{k}\\ &= \frac{{2 \times \left( {760\;{\rm{N}}} \right) \times \left( {1.0\;{\rm{m}}} \right)}}{{12.67 \times {{10}^5}\;{\rm{N/m}}}}\\ &= 120.0 \times {10^{ - 5}}\;{{\rm{m}}^2}\end{aligned}\)

Thus, \(x' = 34.6 \times {10^{ - 3}}\;{\rm{m}}\)

The bathroom scale will show a reading equal to the restoring force in the spring. Thus, the restoring force in the spring when you jump on the bathroom scale from height h is:

\(\begin{aligned}F' &= kx'\\ &= \left( {12.67 \times {{10}^5}\;{\rm{N/m}}} \right) \times \left( {34.6 \times {{10}^{ - 3}}\;{\rm{m}}} \right)\\ &= 4.36 \times {10^4}\;{\rm{N}}\end{aligned}\)

Thus, the bathroom scale reads\(4.36 \times {10^4}\;{\rm{N}}\).