Answer:
The energy stored in the spring when the bike goes over a bump is [tex]6.6\times 10^{-3}[/tex] 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 ([tex]U_{g}[/tex]), measured in joules, is defined by:
[tex]U_{e} = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]x[/tex] - Deformation, measured in meters.
If we know that [tex]k = 33\,\frac{N}{m}[/tex] and [tex]x = 0.02\,m[/tex], the energy stored by the spring due to compression is:
[tex]U_{e} = \frac{1}{2}\cdot \left(33\,\frac{N}{m} \right) \cdot (0.02\,m)^{2}[/tex]
[tex]U_{e} = 6.6\times 10^{-3}\,J[/tex]
The energy stored in the spring when the bike goes over a bump is [tex]6.6\times 10^{-3}[/tex] joules.