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ollegr [7]
4 years ago
12

A mass weighing 30 lbstretches a spring . The mass is also attached to a damper with coefficient . Determine the value of for wh

ich the system is critically damped. Assume that .
Physics
1 answer:
LUCKY_DIMON [66]4 years ago
4 0

Answer:

the damping coefficient when the system is critically damped is 13.42 lb s/ft.

Explanation:

The directions are missing the last part, which is:

"A mass weighing 30 lb stretches a spring 7.5 in. The mass is also attached to a damper with coefficient γ. Determine the value of γ for which the system is critically damped. Assume that g = 32 \cfrac {ft}{s^2}

Round your answer to three decimal places."

The spring system is determined by the following differential equation

my''+\gamma y'+ky=0

where \gamma is the damping coefficient, thus it is critically damped when the system transitions from real to complex solutions that happens at

\gamma_^2 -4km=0

Solving for the damping coefficient.

\gamma^2 = 4km\\ \gamma = \sqrt{4km}

where the spring constant k is given by

k = \cfrac{mg}{L}\\k = \cfrac wL\\

And the mass is given by

mg = w\\ m = \cfrac wg

So the damping coefficient will be

\gamma = \sqrt{4\cfrac wL \cfrac wg}\\\gamma = \sqrt{4\cfrac {w^2}{gL}}

Replacing the given information we have:

\gamma = \sqrt{4\cfrac {(30\, lbs)^2}{32\cfrac{ft}{s^2}0.625 ft}

Thus we get

\gamma = 13.42 \cfrac{lb \cdot s}{ft}

The value of the damping coefficient when the system is critically damped is 13.42 lb s/ft.

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2 years ago
A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi
tekilochka [14]

Answer:

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg

Explanation:

Hi there!

Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:

PE = EPE

m · g · h = 1/2 k · x²

Where:

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h = height.

k = spring constant.

x = compression distance

The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:

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Then, using the equation of gravitational potential energy:

PE = m · g · h =  541.2 J

m =  541.2 J/ g · h

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A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po
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Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5  / x² = 2 / ( .4 - x )²

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