Dr Gulshan S. answered • 12/09/17

PhD In Physics and experience of teaching IB Physics and Math High sc

^{ 2}) = 7 = 1/2 ( K * 0.2*0.2)

Becky T.

asked • 12/08/17Suppose that a spring has a natural length of 20 cm, and that 7 J of work is needed to stretch from a length of 30 cm to 50 cm. How far beyond its natural length will a force of 28 N keep the spring stretched?

I have that the answer is 16 cm.

Thank you

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Dr Gulshan S. answered • 12/09/17

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PhD In Physics and experience of teaching IB Physics and Math High sc

To me it appears that stretching is from 30cm to 50 cm = 20 cm for a work or PE of 7 J

so 1/2(K X^{ 2}) = 7 = 1/2 ( K * 0.2*0.2)

should be used to the value of K = 350 N/m

From here we can go to the second part of question

Arturo O. answered • 12/08/17

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Experienced Physics Teacher for Physics Tutoring

Use the given information to find the spring constant, and then use Hooke's law to find the stretching under the given force. The change in potential energy from 0.30 to 0.50 m is

ΔU = 7 J [given]

Remember to subtract the unstretched length from the final length to get the Δx that you plug into the potential energy.

ΔU = (k/2)[(0.50 - 0.20)^{2} - (0.30 - 0.20)^{2}] = (k/2)(0.08) = 0.04k

7 J = 0.04k ⇒ k = 175 N/m

F = kΔx = 175Δx

F = 28 N ⇒

28 = 175Δx

[Note that it is stretched to a final length of (20 + 16) cm = 36 cm.]

Michael J. answered • 12/08/17

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4+ Experience in 2D and 3D Drafting

Use the formula

PE = (1/2)kΔx^{2}

to find the spring constant. The elongation is 30 cm. We must convert this to meters.

30 cm = 0.3 m

7 = (1/2)k(0.3)^{2}

7 = (1/2)(0.09)k

7 = 0.045k

155 = k

Now we plug in this value into Hooke's law to find the distance that will keep it stretched at 28 N.

F = kx

F / k = x

28 / 155 = x

0.18 = x

0.18 m = 18 cm

The spring must stretched 18 cm beyond its natural height.

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