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WAEC Physics 2016 Paper

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Question : 34

Total: 47

A man and his friend have masses 70kg and 60kg respectively. If they are seated 1.0m apart, calculate the gravitational force of attraction between them. (G = 6.67 x 10^-11 Nm² kg^-2)

1.59 x 10^-14 N
2.80 x 10^-7 N
1.30 x 10^-2 N
6.30 x 10¹³ N

Solution:

Concept:

Newton's law of universal gravitation gives the attractive force between any two masses.

Explanation:

Given:

m_1 = 70\ \text{kg}

,

m_2 = 60\ \text{kg}

,

r = 1.0\ \text{m}

.

G = 6.67 \times 10^{-11}\ \text{N m}^2\text{ kg}^{-2}

.

Use the formula:

F = \dfrac{G m_1 m_2}{r^2}

.

Substitute the values:

F = \dfrac{6.67 \times 10^{-11} \times 70 \times 60}{(1.0)^2}

.

Simplify:

70 \times 60 = 4200

and

(1.0)^2 = 1

.

So

F = 6.67 \times 10^{-11} \times 4200

.

Therefore,

F = 2.8014 \times 10^{-7}\ \text{N}

.

Rounding to two decimal places gives

F = 2.80 \times 10^{-7}\ \text{N}

.

Answer:

The gravitational force of attraction between them is

2.80 \times 10^{-7}\ \text{N}

, which is option B.

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