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

Answer:

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

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

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

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:

$PE_g=PE_s$

$m.g.h=\frac{1}{2} \times k.\Delta x^2$

$1.4\times 9.8\times 1.15=0.5\times 300\times \Delta x^2$

$\Delta x =0.32433\ m= 324.33\ mm$ 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:

$w'=m.g-0.6$

$w'=1.4\times 9.8-0.6$

$w'=1.01\ N$

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

$PE_g'=PE_s$

$w'.h=\frac{1}{2} \times k.\delta x^2$

$1.01\times 1.15=0.5\times 300\times \delta x^2$

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

c)

On moon, as per given details:

$m.g'.h=\frac{1}{2} \times k.\delta x^2$

$1.4\times 1.63\times 1.15=0.5\times 300\times \delta x^2$

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