Question #160914
At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0N/m is set into oscillatory motion with amplitude of 20.0cm. It is observed that the maximum speed of the bunch of bananas is 40.0cm/s. What is the weight of the bananas?
1
Expert's answer
2021-02-10T10:11:27-0500

The maximum speed of the bunch of bananas undergoing simple harmonic motion can be found as follows:


vmax=ωA=kmA.v_{max}=\omega A=\sqrt{\dfrac{k}{m}}A.

Then, from this equation, we can find the mass of the bananas:


m=kA2vmax2.m=\dfrac{kA^2}{v_{max}^2}.

Finally, we can find the weight of the bananas:


W=mg=kA2gvmax2,W=mg=\dfrac{kA^2g}{v_{max}^2},W=16.0 Nm(0.2 m)29.8 ms2(40 cms1 m100 cm)2=39.2 N.W=\dfrac{16.0\ \dfrac{N}{m}\cdot(0.2\ m)^2\cdot9.8\ \dfrac{m}{s^2}}{(40\ \dfrac{cm}{s}\cdot\dfrac{1\ m}{100\ cm})^2}=39.2\ N.

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