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Sagot :
Given,
The spring constant, k=250 N/m
The compression of the spring, x=0.73 m
The mass of the weed, m=1.5 kg
The potential energy of the spring is given by,
[tex]U=\frac{1}{2}kx^2[/tex]According to the law of the conservation of energy, the total energy of the spring and the weed should be conserved. That is when the spring is expanded, it loses its potential energy which is gained by the weed as kinetic energy.
Thus the potential energy of the spring is equal to the kinetic energy of the weed.
The kinetic energy of the weed is given by,
[tex]E_K=\frac{1}{2}mv^2[/tex]Where v is the maximum speed of the weed.
From the law of conservation of energy,
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ \Rightarrow v^2=\frac{kx^2}{m} \\ v=\sqrt[]{\frac{k}{m}}\times x \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{\frac{250}{1.5}}\times0.73 \\ =9.42\text{ m/s} \end{gathered}[/tex]Therefore the maximum speed of the block of weed is 9.42 m/s
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