Question

A 1.40 kg object is held 1.15 m above a relaxed, massless vertical spring with a force constant of 300 N/m. The object is dropped onto the spring.

How far does the object compress the spring?
Repeat part (a), but now assume that a constant air-resistance force of 0.600 N acts on the object during its motion.
How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2 and air resistance is neglected?

Answer 1

a)
`\Delta x =0.32433\ m= 324.33\ mm`

b)
`\delta x=0.087996\ m=87.996\ mm`

c)
`\delta x=0.13227\ m=132.27\ mm`

Step-by-step explanation:

Given:

• mass of the object,
  `m=1.4\ kg`

• height of the object above the spring,
  `h=1.15\ m`

• spring constant,
  `k=300\ N.m^(-1)`

a)

When the object is dropped onto the spring whole of the gravitational potential energy of the mass is converted into the spring potential energy:

`PE_g=PE_s`

`m.g.h=(1)/(2) * k.\Delta x^2`

`1.4* 9.8* 1.15=0.5* 300* \Delta x^2`

`\Delta x =0.32433\ m= 324.33\ mm` is the compression in the spring

b)

When there is a constant air resistance force of 0.6 newton then the apparent weight of the body in the medium will be:

`w'=m.g-0.6`

`w'=1.4* 9.8-0.6`

`w'=1.01\ N`

Now the associated gravitational potential energy is converted into the spring potential energy:

`PE_g'=PE_s`

`w'.h=(1)/(2) * k.\delta x^2`

`1.01* 1.15=0.5* 300* \delta x^2`

`\delta x=0.087996\ m=87.996\ mm`

c)

On moon, as per given details:

`m.g'.h=(1)/(2) * k.\delta x^2`

`1.4* 1.63* 1.15=0.5* 300* \delta x^2`

`\delta x=0.13227\ m=132.27\ mm`