In order for maximum work to be done...
2 answers:
Answer:
For constant traveled distance and applied forces, apply the for parallel to the direction of the movement.
Explanation:
There are 3 components in the amount of work produced when applying a force to an object.
1. The amount of force applied
2. The distance that the object is traveled
3. The angle between the force vector and the distance vector
- To get the maximum amount of work for a constant force and distance, the force needs to be along the distance traveled
- If the angle and the distance are constant, the only way is to apply a larger force in magnitude
- If the angle and the force are constant, increasing the distance traveled will result in higher amount of work produced.
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Quasi frequency = 4√6
Quasi period = π√6/12
t ≈ 0.4045
<u>Explanation:</u>
Given:
Mass, m = 20g
τ = 400 dyn.s/cm
k = 3920
u(0) = 2
u'(0) = 0
General differential equation:
mu" + τu' + ku = 0
Replacing the variables with the known value:
20u" + 400u' + 3920u = 0
Divide each side by 20
u" + 20u' + 196u = 0
Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.
r² + 20r + 196 = 0
Determining the roots:
r = -10 ± 4√6i
The general solution for two complex roots are:
y = c₁ eᵃt cosbt + c₂ eᵃt sinbt
with a the real part of the roots and b be the imaginary part of the roots.
Since, a = -10 and b = 4√6
u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t
u(0) = 2
u'(0) = 0
(b)
Quasi frequency:
μ =
(c)
Quasi period:
T = 2π / μ
(d)
|u(t)| < 0.05 cm
u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05
solving for t:
τ = t ≈ 0.4045
