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grigory [225]
3 years ago
12

To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x

is in meters. What is the change in potential energy ∆U when the spring is stretched 2 meters from its equilibrium position?
Physics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

64 J

Explanation:

The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

=  ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U =  ∫₀²F.dx

=  ∫₀²(40x - 6x²).dx

=  ∫₀²(40xdx - 6x²dx)

=  ∫₀²(40x²/2 - 6x³/3)

=  ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

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kifflom [539]

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A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student
castortr0y [4]

Answer:

5x10^-3

Explanation:

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Hooke's Law can be represented as

<h3> F = kx, </h3>

<em>where F is the force </em>

<em>            k is the spring constant</em>

<em>            x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring =  5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

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