Question

A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released
a distance of 5 m above the platform. When the block is momentarily at rest, the spring is
compressed by 25 cm. Find the speed of the block when the compression of the spring is only
15 cm.

Answer 1

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

• mass of the block, m = 2.4 kg
• height of the block, h =  5 m
• compression of the spring, x = 25 cm = 0.25 m

The spring constant is calculated as follows;

$F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m$

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

$\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2 - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2 = mgh - \frac{1}{2} kx^2\\\\mv^2 = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s$

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

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