Question

An object of inertia 0.5kg is hung from a spring, and causes it to extend 5cm. In an elevator accelerating downward at 2 m/s^2 , how far will the spring extend if the same object is suspended from it? Draw the free body diagrams for both the accelerating and non-accelerating situations.

Answer 1

Answer:3.98cm

Step-by-step explanation:

given data

mass of object
`\left ( m\right )`=0.5kg

intial extension=5 cm

elevator acceleration=2 m/
`s^2`

From FBD of intial position

Kx=mg

K=
`(0.5* 9.81)/(0.05)`

k=98.1 N/m

From FBD of second situation

mg-k
`x_0`=ma

k
`x_0`=m(g-a)

`x_0`=
`(0.5(9.81-2))/(98.1)`

`x_0`=3.98cm