The extension, y, of a material with an applied force, F, is given by y = e^(F x 1 x 10^(-3))(e is to the power of F x 1 x 10; 10 is to the power of -3 in that equation)

Question

a)Calculate the work done if the force increases from 100N to 500N using:

i) An analytical integration technique

ii) A numerical integration technique

[Note: the work done is given by the area under the curve]

Answer 1

It's a little hard to understand this problem because "extension" is not a word typically applied to something like this. I can only assume it means the material moves by the amount "y" when pushed by the applied force and that the "move" is done in the same direction as the applied force.

If y = e^0.001F then ln(y) = 0.001F or F = ln(y)/0.001 or F = 1000ln(y)

When F = 100 we can solve for y:

100 = 1000ln(y)

0.1 = ln(y)

e^0.1 = y

When F = 100 we can solve for y:

500 = 1000ln(y)

0.5 = ln(y)

e^0.5 = y

Work = _start∫^finish F(y)dy so:

You can use a TI-84 to do the numerical integration.

The antiderivative of ln(y) is found using integration by parts:

∫udv = uv - ∫vdu where u = ln(y), du/dy = 1/y, dv/dy = 1, v = y

∫ln(y) dy = ln(y)•y - ∫y•(dy/y)

∫ln(y) dy = yln(y) - ∫dy

∫ln(y) dy = yln(y) - y

so 1000∫ln(y) dy = 1000y(ln(y) - 1)

Therefore the analytical solution is:

1000y(ln(y) - 1) evaluated between e^0.1 and e^0.5 =

1000e^0.5(ln(e^0.5) -1) - 1000e^0.1(ln(e^0.1) - 1)

1000e^0.5(0.5 - 1) - 1000e^0.1(0.1 - 1)

1000e^0.5(-0.5) - 1000e^0.1(-0.9)

-500e^0.5 + 900e^0.1

900e^0.1 - 500e^0.5 ≈ 170.3 Nm