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Question

Answers

a. $ 16.66\;{\rm{N}} $

b. $ 15\;{\rm{N}} $

c.$ 17.53\;{\rm{N}} $

d.$ 16\;{\rm{N}} $

Answer
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9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} $ with $ \dfrac{1}{6} $.

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.

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} $.