Respuesta :
Answer
given,
mass of the = m₁ = 8.75 Kg
another mass of the object = m₂ = 14 Kg
distance between them = 50 cm
R₁ = 17 cm
R₂ = 50 -17 = 33 cm
a) Force applied due to the Mass 8.75 in +ve x- direction
[tex]F_1 = \dfrac{GM_1 m}{R_1^2}[/tex]
[tex]F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}[/tex]
[tex]F_1 = 2.019\times 10^{8}\ m[/tex]
Force applied due to mass 14 Kg in -ve x-direction
[tex]F_2 = \dfrac{GM_2 m}{R_2^2}[/tex]
[tex]F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}[/tex]
[tex]F_2 = 0.857\times 10^{8}\ m[/tex]
net force
F = F₁ + F₂
[tex]F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m[/tex]
[tex]F = 1.162\times 10^{8}\ m [/tex]
Using newton second law
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{ 1.162\times 10^{8}\ m}{m}[/tex]
[tex]a =1.162\times 10^{8} \ m/s^2[/tex]
b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg
(a) The acceleration of the particle is 1.163 x 10⁻⁸ m/s².
(b) The acceleration of the particle is toward the 8.75kg mass.
Force between the masses
The force between the masses is determined by applying Newton's law of universal gravitation as shown below;
F = Gm₁m₂/R²
Force between 8.75 kg and mass, m
[tex]F_1 = \frac{6.67 \times 10^{-11} \times 8.75 \times m}{(0.17)^2} \\\\F_1 = 2.02 \times 10^{-8} m[/tex]
Force between 14 kg and mass, m
distance between 14 kg and m = 50 cm - 17 cm = 33 cm
[tex]F_2 = \frac{6.67 \times 10^{-11} \times 14 \times m}{(0.33)^2} \\\\F_2 = 8.57\times 10^{-9} m= 0.857\times 10^{-8}m[/tex]
Net force
The net force on the masses;
F(net) = F1 + F2
F(net) = 2.02 x 10⁻⁸m - 0.857 x 10⁻⁸m = 1.163 x 10⁻⁸ m
Acceleration of the particle
F = ma
a = F/m
a = (1.163 x 10⁻⁸ m)/m
a = 1.163 x 10⁻⁸ m/s²
Since the acceleration if positive it will be directed towards 8.75 kg mass.
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