skullglitches
24.03.2020 •
Physics
A hydraulic cylinder on an industrial machine pushes a steel block a distance of x feet (0 ≤ x ≤ 6), where the variable force required is F(x) = 1100xe−x pounds. Find the work done in pushing the block the full 6 feet through the machine. (Round your answer to three decimal places.)
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Ответ:
W = 1080.914 J
Explanation:
f(x) = 1100xe⁻ˣ
Work done by a variable force moving through a particular distance
W = ∫ f(x) dx (with the integral evaluated between the interval that the force moves through)
W = ∫⁶₀ 1100xe⁻ˣ dx
W = 1100 ∫⁶₀ xe⁻ˣ dx
But the integral can only be evaluated using integration by parts.
∫ xe⁻ˣ dx
∫ vdu = uv - ∫udv
v = x
(dv/dx) = 1
dv = dx
du = e⁻ˣ dx
∫ du = ∫ e⁻ˣ dx
u = -e⁻ˣ
∫ vdu = uv - ∫udv
∫ xe⁻ˣ dx = (-e⁻ˣ)(x) - ∫ (-e⁻ˣ)(dx)
= -xe⁻ˣ - e⁻ˣ = -e⁻ˣ (x + 1)
∫ xe⁻ˣ dx = -e⁻ˣ (x + 1) + C (where c = constant of integration)
W = 1100 ∫⁶₀ xe⁻ˣ dx
W = 1100 [-e⁻ˣ (x + 1)]⁶₀
W = 1100 [-e⁻⁶ (6 + 1)] - [-e⁰ (0 + 1)]
W = 1100 [-0.0173512652 + 1]
W = 1100 × (0.9826487348)
W = 1080.914 J
Hope this Helps!!!
Ответ:
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