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Hint: In this question consider the gravitational force on the surface of the moon be ${g_m}$ that is ${g_m} = \dfrac{g}{6} = \dfrac{{9.8}}{6}$. Then use that the force acting upon the body is actually the weight of the body and is the product of the mass and acceleration due to gravity that is $F = W = Mg$. This will help approaching the problem.
Complete step-by-step solution -
According to Newton’s second law of motion the force acting on the body is the product of the mass of the body and the acceleration by which the body is moving.
In case if a body is at a rest position the force acting on the body is equal to the weight of the body which is acting downward on the body and this is equal to the product of mass of the body and the acceleration due to gravity.
$ \Rightarrow F = W = Mg$Newton (N)..................... (1)
Where, W = weight of the body, M = mass of the body and g = acceleration due to gravity.
Now it is given that the mass (M) of an object = 10Kg.
And gravitational force on the surface of the moon is (1/6) times the gravitational force on the earth.
As we know on the earth surface the gravitational force, g = 9.8 m/s2.
So on the surface of the moon the gravitational force will be, ${g_m} = \dfrac{g}{6} = \dfrac{{9.8}}{6}$ m/s2.
Now from equation (1) weight of the object on the surface of the earth is,
$ \Rightarrow {W_e} = 10 \times 9.8 = 98N$
And the weight of the object on the surface of the moon is,
$ \Rightarrow {W_m} = 10 \times \dfrac{{9.8}}{6} = 16.66N$
So this is the required weight of the object on the surface of the earth and on the moon respectively.
Note – The main reason behind the lesser gravity on the surface of the moon as compared to the earth is that moon’s size is comparatively very less as compared to the earth moreover moon is also less dense as compared to that of earth. Less dense allows weak force of gravitational attraction on the moon.
Complete step-by-step solution -
According to Newton’s second law of motion the force acting on the body is the product of the mass of the body and the acceleration by which the body is moving.
In case if a body is at a rest position the force acting on the body is equal to the weight of the body which is acting downward on the body and this is equal to the product of mass of the body and the acceleration due to gravity.
$ \Rightarrow F = W = Mg$Newton (N)..................... (1)
Where, W = weight of the body, M = mass of the body and g = acceleration due to gravity.
Now it is given that the mass (M) of an object = 10Kg.
And gravitational force on the surface of the moon is (1/6) times the gravitational force on the earth.
As we know on the earth surface the gravitational force, g = 9.8 m/s2.
So on the surface of the moon the gravitational force will be, ${g_m} = \dfrac{g}{6} = \dfrac{{9.8}}{6}$ m/s2.
Now from equation (1) weight of the object on the surface of the earth is,
$ \Rightarrow {W_e} = 10 \times 9.8 = 98N$
And the weight of the object on the surface of the moon is,
$ \Rightarrow {W_m} = 10 \times \dfrac{{9.8}}{6} = 16.66N$
So this is the required weight of the object on the surface of the earth and on the moon respectively.
Note – The main reason behind the lesser gravity on the surface of the moon as compared to the earth is that moon’s size is comparatively very less as compared to the earth moreover moon is also less dense as compared to that of earth. Less dense allows weak force of gravitational attraction on the moon.
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