Question

1.34 kg object is held 1.49 m above a relaxed, massless vertical spring with a force constant of 309 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 0.720 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

let h be distance through which the spring is compressed.

a)

Initial potential energy of the object is 1.34*9.8*(1.49+h)

This is equal to the potential energy of the compressed spring

1.34*9.8*(1.49+h)= 0.5*309*h²

Solving for h

h = 0.40 m

--------------------------------------...

b)

1.34*9.8*(1.49+h) – 0.72*(1.49+h) = 0.5*309*h²

h = 0.39 m

---------------------------------

c

1.34*9.8*(1.49+h) /1.63 = 0.5*309*h²

1.34*9.8*(1.49+h) = 1.63*0.5*309*h²

h = 0.31 m

============================