Step-by-Step Solution
Step 1
Given that the weight of the man on Earth is $155\;{\rm{lb}}$, and the acceleration due to gravity on the moon is ${g_m} = 5.30\;{\rm{ft/}}{{\rm{s}}^2}$.
We are required to find out the mass of man on Earth in slugs, kilograms and man’s weight in Newton. Also it is required to find out the weight of man on the moon in pounds and mass of man on moon.
Step 2
(a)
The weight $1\;{\rm{lb}}$ is equal to the $0.031\;{\rm{slug}}$.
Step 3
Convert the units.\[\begin{array}{c} 155\;{\rm{lb}} = 155\;{\rm{lb}} \times \left( {\frac{{0.031\;{\rm{slug}}}}{{1\;{\rm{lb}}}}} \right)\\ = 4.80\;{\rm{slugs}} \end{array}\]
Step 4
(b)
The weight $1\;{\rm{lb}}$ is equal to the $0.4536\;{\rm{kg}}$.
Step 5
Convert the units.\[\begin{array}{c} 155\;{\rm{lb}} = 155\;{\rm{lb}} \times \left( {\frac{{0.4536\;{\rm{kg}}}}{{1\;{\rm{lb}}}}} \right)\\ = 70.3\;{\rm{kg}} \end{array}\]
Step 6
(c)
The equation for the weight of man on Earth is given by,\[{W_E} = {m_{\left( {{\rm{kg}}} \right)}} \times {g_E}\]
Here, ${g_E}$ is the acceleration due to gravity on the Earth, having a standard value of $9.81\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$ .
Step 7
Substitute all the values in the above equation.\[\begin{array}{c} {W_E} = 70.3\;{\rm{kg}} \times 9.81\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}\\ = 6.89 \times {10^2}\;\left( {\frac{{{\rm{kg}} \cdot {\rm{m}}}}{{{{\rm{s}}^2}}}} \right)\\ = 6.89 \times {10^2}\;\left( {\frac{{{\rm{kg}} \cdot {\rm{m}}}}{{{{\rm{s}}^2}}}} \right) \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}}}} \right)\\ = 6.89 \times {10^2}\;{\rm{N}} \end{array}\]
Step 8
Now convert the units in kilo-Newton.\[\begin{array}{c} {W_E} = 6.89 \times {10^2}\;{\rm{N}}\\ = 0.689 \times {10^3}\;{\rm{N}} \times \left( {\frac{{1\;{\rm{kN}}}}{{{{10}^3}\;{\rm{N}}}}} \right)\\ = 0.689\;{\rm{kN}} \end{array}\]
Step 9
(d)
The length $1\;{\rm{ft}}$ is equal to the $0.3048\;{\rm{m}}$.
Step 10
The acceleration due to gravity on the moon in ${{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$ can be calculated as,\[\begin{array}{c} {g_m} = 5.30\;{\rm{ft/}}{{\rm{s}}^2}\\ = 5.30\;\left( {\frac{{{\rm{ft}}}}{{{{\rm{s}}^2}}}} \right) \times \left( {\frac{{0.3048\;{\rm{m}}}}{{1\;{\rm{ft}}}}} \right)\\ = 1.61\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}} \end{array}\]
Step 11
The equation for the weight of man on moon is given by,\[{W_m} = {m_{\left( {{\rm{kg}}} \right)}} \times {g_m}\]
Step 12
Substitute all the values in the above equation.\[\begin{array}{c} {W_m} = 70.3\;{\rm{kg}} \times 1.61\;{\rm{m/}}{{\rm{s}}^2}\\ = 1.13 \times {10^2}\;\left( {{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}} \right) \times \left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{1\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}}}} \right)\\ = 1.13 \times {10^2}\;{\rm{N}} \times \left( {\frac{{0.2248\;{\rm{lb}}}}{{1\;{\rm{N}}}}} \right)\\ = 25.4\;{\rm{lb}} \end{array}\]
Step 13
(e)
The value of mass remains constant and do not vary with the gravitational acceleration of the place. Therefore, the mass of the man on the moon is $70.3\;{\rm{kg}}$.