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When a mass m1is placed on top of a spring and allowed to come to rest, the spring iscompressed by an amount x1. How much would the spring compress if it is an elevatoraccelerating upwards at 3g and has a mass of 4m1placed on top of the spring?

Respuesta :

Answer:

 x₂ = 16 g m₂ / k

Explanation:

The spring in all cases must comply with Hooke's law

              F = k x

Newton's diagram for equilibrium has

             F - W = 0

             k x₁ = m₁ g

             k = g m₁ / x₁

When the elevator moves you have a clarification

            F-W = m a

            F = m (g + a)

Compression is

            K x₂ =4 m (g + 3g)

               x₂ = 4m / k (4g)

                x₂ = 16m2 / k g

When the spring compress If it is an elevator accelerating upwards at 3g and has a mass of 4m1placed on top of the spring is = x₂ = 16 g m₂ / k

Calculation of Elevator Accelerating

When The spring in all cases Then it must comply with Hooke's law

After that, F = k x

Then Newton's diagram for equilibrium has

Then F - W = 0

Now, k x₁ = m₁ g

Hence, k = g m₁ / x₁

When the elevator moves you have a clarification mention in below:

Then F-W = m a

F = m (g + a)

Now, Compression is

Then K x₂ =4 m (g + 3g)

x₂ = 4m / k (4g)

Therefore, x₂ = 16m2 / k g

Find more information about Elevator Accelerating here:

https://brainly.com/question/6080367

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