A 0.250-kg block is placed on a light vertical spring $k = 5.00 \times 10^3 \text{ N/m}$ and pushed downwards, compressing the spring 0.100 m.

A 0.250-kg block is placed on a light vertical spring $k = 5.00 \times 10^3 \text{ N/m}$ and pushed downwards, compressing the spring 0.100 m.


Asked by 3 years ago
393k points

A 0.250-kg block is placed on a light vertical spring $k = 5.00 \times 10^3 \text{ N/m}$ and pushed downwards, compressing the spring 0.100 m.  After the block is released, it leaves the spring and continues to travel upwards.  What height above the point of release will the block reach if air resistance is negligible?

A 0
bhdrkn

1 Answer

Answered by 3 years ago
389.5k points

The potential energy contained in a spring is:

$ PE_s = \frac{1}{2}k x^2 $

The potential energy of the spring at its lowest point, is turned into gravitational potential energy at it's highest point.  Therefore,

$ \frac{1}{2}k x^2 = mgh \,\,\, \to \,\,\, h=\frac{k x^2}{2gm} = \frac{(5.00 \times 10^3 \, N/m)(0.100 \, m)}{2(9.8 \, m/s^2)(0.250 \, kg)} = 102 \, meters $

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Asked: 3 years ago

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