Question

A 1.39 kg object is held 1.29 m above a relaxed, massless vertical spring with a force constant of 294 N/m. The object is dropped onto the spring.

(a) How far does the object compress the spring?


(b) Repeat part (a), but now assume that a constant air-resistance force of 0.768 N acts on the object during its motion.


(c) 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

As we know that;

a)

The initial energy is potential energy:

= Ue= mgh

= 1.39kg*9.81m/s^2*1.29m

= 17.59J

So, the final energy is the spring energy,

= Pe= 1/2kx^2

Then,

The initial energy must equal the final energy,

= Ue=Pe

= x = ?(17.59J*2 / 294 N/m) = 0.345 m

b)

The work done by the air is equal to:

= W = F*d

= 0.768N*1.29m

= 0.9907 J

Therefore,

The energy remaining before it gets to the springs,

= 17.59 J - 0.9907J

= 16.599J

= x = ?(16.599J *2 / 294N/m) = 0.336 m

Then,

c)

If on the moon:

= Ue= mgh

= 1.39kg*1.63m/s^2*1.29m

= 2.92J

= x = ?(2.92J*2 /294N/m)

= 0.14m