A spring has a natural length of 16 cm. If a 22-N force is required to keep it stretched to a length of 26 cm, how much work W is required to stretch it from 16 cm to 21 cm? (Round your answer to two decimal places.)
I need getting a number less than which is obviously not the answer. And I need to put the answer is Joule form?
I need getting a number less than which is obviously not the answer. And I need to put the answer is Joule form?
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F = k(x - 0.16) . . . where x is the length in metres.
22 = k(0.26 - 0.16)
22 = 0.1k
k = 220
F = 220(x - 0.16)
W = ∫ F dx
= ∫ 220(x - 0.16) dx
= ∫ (220x - 35.2) dx
= [ 110x² - 35.2x ] from x = 0.16 to 0.21
= (4.851 - 7.392) - (2.816 - 5.632)
= -2.541 - -2.816
= 0.275 Joule
22 = k(0.26 - 0.16)
22 = 0.1k
k = 220
F = 220(x - 0.16)
W = ∫ F dx
= ∫ 220(x - 0.16) dx
= ∫ (220x - 35.2) dx
= [ 110x² - 35.2x ] from x = 0.16 to 0.21
= (4.851 - 7.392) - (2.816 - 5.632)
= -2.541 - -2.816
= 0.275 Joule
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The first part of the statement will allow you to determine the coefficient in Hooke's law (spring constant).
22N = k10cm = 0.1km ==> k = 220 N/m.
The variable force F(x) = 220x N where x is the displacement from equilibrium in meters. The work sought is
0.05
∫ 220x dx = 0.275 Nm
0
This value is exact. If you really want two decimal digits, the work W ≈ 0.28 J. A Newton-meter is a Joule.
22N = k10cm = 0.1km ==> k = 220 N/m.
The variable force F(x) = 220x N where x is the displacement from equilibrium in meters. The work sought is
0.05
∫ 220x dx = 0.275 Nm
0
This value is exact. If you really want two decimal digits, the work W ≈ 0.28 J. A Newton-meter is a Joule.