A spring scale being used to measure the weight of an object reads 17.1 N when it is used on earth. The spring stretches 4.40 cm under the load. The same object is weighed on the moon, where gravitational acceleration is

1 /6 g.

Find the reading of the spring scale on the moon.

Find the period for vertical oscillations of the spring on the moon.

This question has been really bugging me! Can someone please provide the work and answer so I can work backwards to learn this!? THANK YOU VERY MUCH FOR YOUR TIME !!!

on earth we are given that

W = m g = 9.81 m
= 17.1 N
so m = 17.1/9.81 = 1.74 kg

F = m a

in static case a = 0 so
m g - k x = 0
x = m g/k
.044 = 17.1/k
so
k = 389 Newtons/meter

on moon
Gmoon = g/6 = 9.81/6 = 1.64 m/s^2
mg moon = (1/6) mg earth = 17.1/6
stretch moon = stretch earth/6 = 4.4/6 = .733 cm

for period, we have m and g moon and k
m = 1.74 kg
g = 1.64 m/s^2
k = 389 N/m

for spring and mass
if
x = sin 2 pi t/T
v = (2 pi/T) cos 2 pi t/T
a = - (2 pi/T)^2 sin 2 pi t/T = -(2 pi/T)^2 x

F = - k x = m a
-k x = -m (2 pi/T)^2 x
so
(2 pi/T)^2 = k/m
2 pi/T = sqrt (k/m)
T = 2 pi sqrt (m/k)
so
T on moon = 2 pi sqrt(1.74/389)

By the way, k and m are the same on earth and moon. The period of the spring/mass system has nothing to do with g and is the same on earth and moon. That would NOT be true for a pendulum.

To find the reading of the spring scale on the moon, we need to first determine the force acting on the object on the moon.

The weight of an object is given by the formula:

Weight = mass x gravitational acceleration

On Earth, the gravitational acceleration is approximately 9.8 m/s^2, while on the moon it is 1/6 times the Earth's value, which means 1/6 x 9.8 m/s^2 = 1.6333... m/s^2.

Now, let's calculate the mass of the object. We know that the weight on Earth is 17.1 N, so we can rearrange the formula to solve for mass:

Mass = Weight / Gravitational acceleration

Substituting the values:

Mass = 17.1 N / 9.8 m/s^2 = 1.74487... kg

Now that we know the mass of the object, we can calculate the force acting on it on the moon using the mass and moon's gravitational acceleration:

Force on the moon = Mass x Moon's gravitational acceleration = 1.74487... kg x 1.6333... m/s^2 = 2.85 N (rounded to two decimal places)

Therefore, the reading of the spring scale on the moon is approximately 2.85 N.

Moving on to the second part of the question, to find the period for vertical oscillations of the spring on the moon, we can use the formula for the period of a mass-spring system:

Period (T) = 2π √(m / k)

where m is the mass and k is the spring constant.

Since the mass of the object remains the same, we only need to determine the spring constant on the moon in order to calculate the period.

The spring constant (k) is defined as the force required to stretch or compress a spring by a certain distance. In this case, we need to find the spring constant on the moon.

Using the data given, the spring stretches 4.40 cm under the load on Earth, but we need to convert this to meters for consistency:

Stretched length = 4.40 cm = 0.044 m

Now, we can use Hooke's Law to find the spring constant:

Force = Spring constant x Stretched length

Rearranging the formula:

Spring constant = Force / Stretched length

On Earth, the force is 17.1 N and the stretched length is 0.044 m:

Spring constant on Earth = 17.1 N / 0.044 m = 388.636...

Now, since the spring constant depends on the material of the spring and is independent of the gravitational acceleration, it remains the same on the moon.

Therefore, the period for vertical oscillations of the spring on the moon will be the same as it was on Earth:

Period (T) = 2π √(m / k) = 2π √(1.74487... / 388.636...) = 0.909... seconds

Therefore, the period for vertical oscillations of the spring on the moon is approximately 0.91 seconds (rounded to two decimal places).

I hope this explanation helps! Let me know if you have any further questions.