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If a spring has a constant of #4 (kg)/s^2#, how much work will it take to extend the spring by #65 cm #?

Answer 1

Leo Abeyta

The work done is #0.845# joules.

We employ the spring's work equation, which says that,

#W=1/2kx^2#

#k# is the spring constant in #"N/m"#

#x# is the extension of the spring in meters

Take note of this:

#1 \ "N/m"=("kg m""/s"^2)/("m")=1 \ "kg/s"^2#

#:.4 \ "kg/s"^2=4 \ "N/m"#

Thus, we have:

#65 \ "cm"=0.65 \ "m"#

Thus, the completed work is:

#W=1/2*4 \ "N/m"*(0.65 \ "m")^2#

#=2 \ "N/m"*0.4225 \ "m"^2#

#=0.845 \ "N m"#

#=0.845 \ "J"#

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Answer 2

Isabella Alvarez

To calculate the work done to extend the spring, you can use the formula:

[ W = \frac{1}{2} k x^2 ]

Where:

( W ) is the work done,
( k ) is the spring constant (4 kg/s² in this case),
( x ) is the displacement from the equilibrium position (0.65 m in this case).

[ W = \frac{1}{2} \times 4 \times (0.65)^2 = 0.845 , \text{J} ]

So, it will take approximately 0.845 joules of work to extend the spring by 65 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.