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?
Δl=10cm=0.1m\Delta l =10cm=0.1mΔl=10cm=0.1m
Ep=4 JE_p=4\ JEp=4 J
Ep=kΔl22E_p=\frac{k\Delta l^2}{2}Ep=2kΔl2
k=2EpΔl2=2∗40.12=0.08 Nmk= \frac{2E_p}{\Delta l^2}=\frac{2*4}{0.1^2}=0.08\ \frac{N}{m}k=Δl22Ep=0.122∗4=0.08 mN
Answer: k=0.08 Nm\text{Answer: }k=0.08\ \frac{N}{m}Answer: k=0.08 mN
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