If a spring has a constant of #4 (kg)/s^2#, how much work will it take to extend the spring by #19 cm #?

Answer 1

3.04J of work.

19 cm is equal to 0.19 m.

Hooke's law says that #FpropDeltax#, or #F=kDeltax#. In this question we know both the extension, and spring constant, so we just have to do 0.19x4=0.76N.

The question however asks for work done, which is represented by #W=FDeltas#, #Deltas=Deltax#. We just need to do 0.76x4=3.04J of work.

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Answer 2

Elena Albarran

The work done to extend the spring by 19 cm can be calculated using the formula: ( W = \frac{1}{2}kx^2 ), where ( W ) is the work done, ( k ) is the spring constant (4 kg/s^2), and ( x ) is the displacement (19 cm or 0.19 m). Plugging in the values, we get: ( W = \frac{1}{2} \times 4 \times (0.19)^2 ). Solving this equation yields the work done to be approximately 0.1448 joules.

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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.