Question

A spacecraft component occupies a volume of 8ft^3 and weighs 25 lb at a location where the acceleration of gravity is 31.0 ft/s^2. Determine its weight, in pounds, and its average density, in lbm/ft^3, on the moon, where g=5.57 ft/s^2.

Answer 1

density=3.125pounds/ft^3

weight=4.35lbf

Step-by-step explanation:

Density is a property of matter that indicates how much mass a body has in a given volume.

It is given by the following equation.

ρ=m/v (1)

where

ρ=density

m=mass

v=volume

The weight on the other hand is the force which the earth (or the moon) attracts to a body with mass, this force is given by the following equation

W=mg (2)

W=weight

m=mass

g=gravity

to solve this problem we have to calculate the mass of the component using the ecuation number 2

W=mg

m=w/g

w=25lbf=804.35pound .ft/s^2

g=32.2ft/s^2

m=804.35/32.2=25 pounds

density

ρ=m/v (1)

ρ=25pounds/8ft^3

ρ=3.125pounds/ft^3 =density

weight in the moon

W=mg

W=(25pounds)(5.57ft/s^2.)=139.25pound .ft/s^2=4.35lbf

Answer 2

The weight is 4.492 lb

The density is
`0.5614 lbm/ft^(3)`

Solution:

As per the question:

Volume of spacecraft component,
`V_(s) = 8ft^(3)`

Mass of the component of spacecraft,
`m_(s) = 25 lb`

Acceleration of gravity at a point on Earth,
`g_(E) = 31.0 ft/s^(2)`

Acceleration of gravity on Moon,
`g_(M) = 5.57 ft/s^(2)`

Now,

The weight of the component,
`w_(c) = m_(c)* (g_(M))/(g_(E))`

`w_(c) = 25* (5.57)/(31.0)`

`w_(c) = 4.492 lb`

Now,

Average density,
`\rho_(avg)`

`\rho_(avg) = \farc{w_(s)}{V_(s)}`

`\rho_(avg) = \farc{4.492}{8} = 0.5614 lbm/ft^(3)`