Answer
Verified
394.5k+ views
Hint: Formula for weight is, $ W = m{g_e} $. Acceleration due to gravity in the surface of the earth, $ {g_e} = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} $. To find the acceleration due to gravity in the surface of the moon, we multiply $ {g_e} =
9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} $ with $ \dfrac{1}{6} $.
Complete step by step answer:
Given, gravitational force on the surface of the moon is only $ \dfrac{1}{6} $ as strong as gravitational force on the earth.
Mass of the object, $ m = 10\;{\rm{kg}} $
Step I:
We know that, acceleration due to gravity in the surface of the earth, $ {g_e} = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} $.
Formula for weight is, $ W = m{g_e} $
Therefore, the weight of the object on earth is, $ W = m{g_e} $.
Substitute the values of $ m $ and $ {g_e} $ in the above equation.
Now, weight of the object on earth is,
$ \begin{array}{c}W = m{g_e}\\ = 10 \times 9.8\\
= 98\;{\rm{N}}\end{array} $
Step II:
Again, according to the question,
Acceleration due to gravity at moon,
$ \begin{array}{c}{g_m} = \dfrac{1}{6} \times 9.8\\ = 1.63\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array} $
Therefore, the weight of the object on the moon is, $ W = m{g_m} $.
Substitute the values of $ m $ and $ {g_m} $ in the above equation.
Now, weight of the object on moon is,
$ \begin{array}{c}W = m{g_m}\\ = 10 \times 1.63\\ =
16.3\;{\rm{N}}\end{array} $
Hence, an object having a mass of $ 10\;{\rm{kg}} $ has the weights on the earth and the moon is $ 98\;{\rm{N}} $ and $ 16.3\;{\rm{N}} $ respectively.
Note:
Know the difference between mass and weight.
Mass: Mass is both a real body component, and a measurement of its acceleration resistance whenever a net force is applied. Kilogram is the fundamental SI unit of mass.
Weight: The weight of an object is the force of gravity acting on the body. Its SI units are
$ {\rm{kg}}{\rm{.m/}}{{\rm{s}}^{\rm{2}}} $ or Newton ($ {\rm{N}} $).
In step I, determine the weight of the object on the earth surface by multiplying the mass of the object to the acceleration due to gravity on earth.
In step II, determine the weight of the object on the moon surface by multiplying the mass of the object to the acceleration due to gravity on earth $ \times \dfrac{1}{6} $.
9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} $ with $ \dfrac{1}{6} $.
Complete step by step answer:
Given, gravitational force on the surface of the moon is only $ \dfrac{1}{6} $ as strong as gravitational force on the earth.
Mass of the object, $ m = 10\;{\rm{kg}} $
Step I:
We know that, acceleration due to gravity in the surface of the earth, $ {g_e} = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} $.
Formula for weight is, $ W = m{g_e} $
Therefore, the weight of the object on earth is, $ W = m{g_e} $.
Substitute the values of $ m $ and $ {g_e} $ in the above equation.
Now, weight of the object on earth is,
$ \begin{array}{c}W = m{g_e}\\ = 10 \times 9.8\\
= 98\;{\rm{N}}\end{array} $
Step II:
Again, according to the question,
Acceleration due to gravity at moon,
$ \begin{array}{c}{g_m} = \dfrac{1}{6} \times 9.8\\ = 1.63\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array} $
Therefore, the weight of the object on the moon is, $ W = m{g_m} $.
Substitute the values of $ m $ and $ {g_m} $ in the above equation.
Now, weight of the object on moon is,
$ \begin{array}{c}W = m{g_m}\\ = 10 \times 1.63\\ =
16.3\;{\rm{N}}\end{array} $
Hence, an object having a mass of $ 10\;{\rm{kg}} $ has the weights on the earth and the moon is $ 98\;{\rm{N}} $ and $ 16.3\;{\rm{N}} $ respectively.
Note:
Know the difference between mass and weight.
Mass: Mass is both a real body component, and a measurement of its acceleration resistance whenever a net force is applied. Kilogram is the fundamental SI unit of mass.
Weight: The weight of an object is the force of gravity acting on the body. Its SI units are
$ {\rm{kg}}{\rm{.m/}}{{\rm{s}}^{\rm{2}}} $ or Newton ($ {\rm{N}} $).
In step I, determine the weight of the object on the earth surface by multiplying the mass of the object to the acceleration due to gravity on earth.
In step II, determine the weight of the object on the moon surface by multiplying the mass of the object to the acceleration due to gravity on earth $ \times \dfrac{1}{6} $.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Define absolute refractive index of a medium
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE