Question #205671

A heavy block is suspended from a vertical spring.

The elastic potential energy is stored in the spring is 4 J.

What is the spring constant if the elongation of the spring is 10 cm?


1
Expert's answer
2021-06-11T11:22:15-0400

Δl=10cm=0.1m\Delta l =10cm=0.1m

Ep=4 JE_p=4\ J

Ep=kΔl22E_p=\frac{k\Delta l^2}{2}

k=2EpΔl2=240.12=0.08 Nmk= \frac{2E_p}{\Delta l^2}=\frac{2*4}{0.1^2}=0.08\ \frac{N}{m}

Answer: k=0.08 Nm\text{Answer: }k=0.08\ \frac{N}{m}



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