Question

What is the force of attraction due to an object of mass 18 kg and
another object of mass 10kg, if the distance between them is 30cm. (with steps)

Answer 1

GRAVITATIONAL FORCE

Given

• Masses of the Objects (m₁,m₂) = 18 kg and 10 kg
• Distance (d) = 30 cm
• Gravitational Constant (G) = 6.674 x 10⁻¹¹ m³
• Force of Attraction (Fg) = ?

The force of attraction can be also called the gravitational forces between two objects.

Convert the distance from cm to m

1 m = 100 cm

(30 cm) / (100 cm / 1 m)

= 0.3 m

Use the formula of Newton's Law of Universal Gravitation to find the force of attraction.

Fg = Gm₁m₂ / d²

Fg = (6.674 x 10⁻¹¹ m³)(18 kg)(10 kg) / (kg • s²)(0.3m)²

Fg = (6.674 x 10⁻¹¹ m³)(18 kg)(10 kg) / (kg • s²)(0.09 m²)

Fg = (6.674 x 10⁻¹¹ m³)(180 kg²) / (kg • s²)(0.09 m²)

Fg = (1.201 x 10⁻⁸ m³ kg²) / (kg • s²)(0.09 m²)

Fg = (1.334 x 10⁻⁷ m³ kg²) / (kg • s²) (m²)

Fg = 1.334 x 10⁻⁷ m • kg/s²

Fg = 1.334 x 10⁻⁷ N

$\:$

Hope it Helps

Answer 2

Mass of one object, $\sf m_1$ = 18 kg

Mass of another object, $\sf m_2$ = 10 kg

Distance between the object, r = 30 cm = 0.3 m

Force of attraction between them = $\sf F_g$

According to Newton's Second Law, $\sf F_g = G \dfrac{m_1 m_2}{r^2}$ ;

$\sf F_g = 6.674 \; × \; 10^{-11} \; m^3 \; kg^{-1} \;s^{-2} × \dfrac{18 \; kg × 10\; kg}{(0.3 \;m)^2}$

• $\sf F_g = \dfrac {1201.32 \; m^3 \; kg \;s^{-2}}{0.09 \;m^2}$

• $\sf F_g = \dfrac {1201.32\; × \;10^{-11} \;m^3 \; kg \;s^{-2}}{0.09 \;m^2}$

• $\sf F_g = 13,348 \;× \;10^{-11} \;m-kg\; s^{-2}$

• $\sf F_g = 1.3348 \;× \;10^{-7} \;m-kg \;s^{-2}$

• $\sf F_g = 1.3348 \;× \;10^{-7} \; N$