Question

A 0.50-kg block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 0.12J. What is the greatest speed of the block?

Answer 1

Given that,

Mass of the block, m = 0.5 kg

Spring constant of the spring, k = 80 N/m

The total mechanical energy is 0.12 J.

To find,

The greatest speed of the block.

Solution,

Total energy of the spring in the spring is given by :

$E=\dfrac{1}{2}kA^2$

A is amplitude

k is spring constant

So,

$A=\sqrt{\dfrac{2k}{E}} \\\\A=\sqrt{\dfrac{2\times 0.21}{80}} \\\\A=0.0724\ m$

Angular frequency is given by :

$\omega=\sqrt{\dfrac{k}{m}} \\\\\omega=\sqrt{\dfrac{80}{0.5}}\\\\\omega=12.64\ rad/s$

The maximum speed of the block is given by :

$v=\omega A\\\\v=12.64\times 0.0724\\\\v=0.91\ m/s$    

So, the greatest speed of the block is 0.91 m/s.

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