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  • A spring hanging from the ceiling of an elevator has a spring constant of 60N/m and a block attached to the other end with a mass of 5kg. If the elevator is accelerating upward at a rate of 3m/s2 a

A spring hanging from the ceiling of an elevator has a spring constant of 60N/m and a block attached to the other end with a mass of 5kg. If the elevator is accelerating upward at a rate of 3m/s2 and the spring is in equilibrium, what is the displacement of the spring? (in meters)

Option: 1

1.08


Option: 2

0.67


Option: 3

2.24


Option: 4

1.54


Answers (1)

best_answer

Solution :

F_{sp}=-k\Delta x

where,

Fsp=spring force

k= spring constant

Δx=net elongation or compression in the spring

By using this concept, let us solve this problem - 

Since the displacement of the spring is at equilibrium, we can write:

F_{net}=0

There are three forces we can account for: spring force, gravitational force, and the additional force resulting from the acceleration of the elevator. If we assume that forces pointing upward are positive, we can write:

F_{spring}-F_{g}-F_{e}=0

Substituting expressions for each force, we get:

kx-mg-ma=0

Rearrange to solve for displacement:

x=\frac{m(g+a)}{k}=\frac{5(10+3)}{60}=1.08 m

 

 

Posted by

Ritika Kankaria

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