Question

10. A mountain bike's

Answer 1

Answer:

The energy stored in the spring when the bike goes over a bump is $6.6\times 10^{-3}$ joules.

Explanation:

Let suppose that spring has a linear behavious, by means of Hooke's Law, definition of Work and Work-Energy Theorem we find that the potential energy stored in the spring ($U_{g}$), measured in joules, is defined by:

$U_{e} = \frac{1}{2}\cdot k\cdot x^{2}$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$x$ - Deformation, measured in meters.

If we know that $k = 33\,\frac{N}{m}$ and $x = 0.02\,m$, the energy stored by the spring due to compression is:

$U_{e} = \frac{1}{2}\cdot \left(33\,\frac{N}{m} \right) \cdot (0.02\,m)^{2}$

$U_{e} = 6.6\times 10^{-3}\,J$

The energy stored in the spring when the bike goes over a bump is $6.6\times 10^{-3}$ joules.