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2 years ago

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A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically eq

ual to the instantaneous velocity. The mass is initially released from a point foot above the equilibrium position with a downward velocity of ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant

1 answer:

andreyandreev [35.5K]2 years ago

6 0

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c) 3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 , spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity hence β = 1

a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>

time mass passes through equilibrium = 1/12 seconds = 0.083

<u>b) Calculate the time at which the mass attains its extreme displacement </u>

time when mass attains extreme displacement = 1/3 seconds = 0.33 secs

<u>c) What is the position of the mass at this instant</u>

position = 3e^-4/3

attached below is the detailed solution to the given problem

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One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:

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The equilibrium condition allows finding the results for the forces of the system are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

N = N₁ + N₂ + N₃ + N₄₄

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N - $W_m -w_{table}$ = 0

N = $W_m +w_{table}$

N = $M_m g + w_{table}$

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

N = 65 9.8 + 100

N = 737 N

c) To calculate the mass of the table we use the relation

W = $m_{table}$ g

$m_{table} = \frac{w_{table}}{g}$

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$m_{table}$ e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Learn more here: brainly.com/question/19860811

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