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If a spring has a constant of #9 (kg)/s^2#, how much work will it take to extend the spring by #22 cm #?

Answer 1
The work done in extending the spring a distance, e, is given by the area under the force-extension graph for the spring. This equals: #W=1/2*F*e# for a spring that obeys Hooke's law.
The relationship between force and extension is given by #F=k*e# If you substitute the above into the first equation you arrive at: #W=1/2*k*e^2# Hence: #W= 1/2*9*0.22^2=0.22J#
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Answer 2

To calculate the work required to extend the spring by 22 cm, we use the formula:

[ \text{Work} = \frac{1}{2}kx^2 ]

where ( k ) is the spring constant and ( x ) is the displacement from the equilibrium position.

First, we convert 22 cm to meters: ( 22 , \text{cm} = 0.22 , \text{m} ).

Then, we plug in the values:

[ \text{Work} = \frac{1}{2} \times 9 , \text{(kg/s}^2) \times (0.22 , \text{m})^2 ]

[ \text{Work} = \frac{1}{2} \times 9 \times 0.0484 ]

[ \text{Work} = 2.178 , \text{J} ]

So, it will take 2.178 J of work to extend the spring by 22 cm.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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