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2 years ago

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A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m

. When the object is a distance 1.25×10−2 m from its equilibrium position, it is observed to have a speed of 0.305 m/s . Part A What is the total energy of the object at any point of its motion

1 answer:

zzz [600]2 years ago

4 0

Answer:

$310.38\times 10^{-4}j$

Explanation:

We have given mass m=0.155 kg

Force constant K = 305 N/m

Distance $X=1.25\times 10^{-2}m$

Velocity $v=.305 m/sec$

The total energy at any position of the motion is give by $E=\frac{1}{2}mv^2+KX^2$ here $\frac{1}{2}mv^2$ is energy due to motion and $KX^2$ is energy due to spring elongation

So total energy $E=\frac{1}{2}\times 0.155\times 0.305^2+305\times 0.0125^2=310.38\times 10^{-4}j$

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