Question

The mass of a boy is 40 kg. What is his weight on the surface of the earth and what is his weight on the surface of the moon? Given that g_e = 9.8 ms^-2 and g_m = $\frac{1}{6}$g_e.

Answer 1

Given:

The mass of a boy (m) = 40 kg.

Acceleration due to gravity (g_e) on the earth = 9.8 ms^-2.

Acceleration due to gravity (g_m) on the moon = $\frac{1}{6}$g_e.

Required to find:

The weight of the boy on the earth (W_e) and also on the moon (W_m).

Solution:

The weight of an object on the earth (W_e) is given by $W_e=m{g}_e$.

By substituting the given values in above equation,

$$\begin{gathered} W_e=m{g}_e \\ \Rightarrow W_e=(40kg)x(9.8m{s}^{-2})=392N \end{gathered}$$

Similarly, the weight of the boy on the moon (W_m) is given by $W_m=m\frac{g_e}{6}$.

$\Rightarrow W_m=(40kg)x(9.8m{s}^{-2})\frac{1}{6}=65.3N$

Therefore, the weight of the boy on the earth is 392 N and on the moon is 65.3 N.