An instrument to measure the acceleration of gravity on Mars is constructed of a spring from which is suspended a mass of 0.40 kg. At a place on earth where the local acceleration of gravity is the spring extends 1.08 cm. When the instrument package is landed on Mars, it radios the information that the spring is extended 0.40 cm. What is the Martian acceleration of gravity?

Question

What will be an ideal response?

Answer 1

The spring constant can be determined from a force balance on the mass on earth. The gravitational force is

The spring force is F = kx, where k is the spring constant and x is the distance it is extended from its equilibrium length. These forces must be equal, so we have

kx = 3.924 N = k * 1.08 cm = k * 0.0108 m.

So, k = 3.924/0.0108 = 363.3 N/m.

Now, we do the same force balance on Mars, with g (for Mars) as the unknown.

F = kx = 363.3 N/m * 0.40 cm = 363.3 N/m * 0.004 m = 1.453 N.

and F = 1.453 N = mg = 0.40 kg * g

So,