When a spring is compressed 2.50 × 10^–2 meter

1 answer:

4 0

The working of a spring is given by:

$T_{el}= \frac{k\Delta x^2}{2}$

Entering the unknowns, we have:

$T_{el}=\frac{k\Delta x^2}{2} \\ 1.25*10^{-2}*2*J=k(2.5*10^{-2}m)^2 \\ k= \frac{2.5*10^{-2}*J}{(2.5*10^{-2}m)^2} \\ k=40* \frac{J}{m^2} \\ k=40 \frac{N*m}{m^2} \\ \boxed {k=40* \frac{N}{m} }$

If you notice any mistake in my english, please let me know, because i am not native.

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A particular automotive wheel has an angular moment of inertia of 12 kg*m^2, and is decelerated from 135 rpm to 0 rpm in 8 secon

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Answer:

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Explanation:

We have given moment of inertia $I=12kgm^2$

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135 rpm = $135\times \frac{2\pi }{60}=14.1371rad/sec$

Time t = 8 sec

So angular speed $\omega _i=135rpm$ and $\omega _f=0rpm$

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-14.1371}{8}=--1.7671rad/sec^2$

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