A body weighs 1000 lbf when exposed to a standard earth gravity g 32.174 ft/s2 . (a) What is its mass in kg? (b) What will the weight of this body be in N if it is exposed to the moon’s standard acceleration gmoon 1.62 m/s2 ? (c) How fast will the body accelerate if a net force of 400 lbf is applied to it on the moon or on the earth?

Question

A body weighs 1000 lbf when exposed to a standard earth gravity [latex]g = 32.174 ft/s^2[/latex] . (a) What is its mass in kg? (b) What will the weight of this body be in N if it is exposed to the moon’s standard acceleration [latex]g_{moon} = 1.62 m/s^2[/latex] ? (c) How fast will the body accelerate if a net force of 400 lbf is applied to it on the moon or on the earth?

Accepted Answer

We need to find the (a) mass; (b) weight on the moon; and (c) acceleration of this body. This is a fairly simple example of conversion factors for differing unit systems. No property data is needed. The example is too low-level for a sketch.
Part (a) Newton’s law (1.2)
F=ma
holds with known weight and gravitational acceleration. Solve for m:
[latex]F= W= 1000 lbf = mg (m)(32.174 ft/s^2)[/latex] , or [latex]m= \frac{1000 lb}{32.174 ft/s^2} =31.08 slugs[/latex]
Convert this to kilograms:
m = 31.08 slugs = (31.08 slugs)(14.5939 kg/slug) =454 kg . (a)
Part (b) : The mass of the body remains 454 kg regardless of its location. Equation (1.2) applies with a new gravitational acceleration and hence a new weight:
[latex]F = w_{moon} = mg_{moon} = (454 kg)(1.62 m/s^2) = 735 N [/latex] . (b)
Part (c) : This part does not involve weight or gravity or location. It is simply an application of
Newton’s law with a known mass and known force:
F= 400 lbf = ma = (31.08 slugs) a
Solve for
[latex]a= \frac{400 lbf} {31.08 slugs} =12.87 \frac{ft} {s^2} (a0.3048 \frac{m} {ft}) = 3.92 \frac{m} {s^2}[/latex] . (c)
Comment (c): This acceleration would be the same on the earth or moon or anywhere.

Question 1.1: A body weighs 1000 lbf when exposed to a standard earth grav...

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