Respuesta :
Answer:[tex]a=2.23\times 10^{-9} m/s^2[/tex]
Explanation:
Given
Mass of first Point mass [tex]m_1=8 kg[/tex]
Mass of second Point [tex]m_2=15 kg[/tex]
distance between them [tex]d=50 cm[/tex]
third point mass [tex]m_3=m[/tex]
Distance between [tex]m\ and\ m_1 is\ 20 cm [/tex]
Distance between [tex]m\ and\ m_2 is 30 cm[/tex]
Force Due to [tex]m_1\ and\ m F_1=\frac{Gmm_1}{d_1^2}[/tex]
[tex]F_1=\frac{8Gm}{(0.2)^2}[/tex]
[tex]F_1=200 mG[/tex]
[tex]F_2=m\frac{Gmm_2}{d_2^2}[/tex]
[tex]F_1=m\frac{15Gm}{(0.3)^2}[/tex]
[tex]F_2=166.67 mG[/tex]
Net Force
[tex]F_{net}=F_1-F_2[/tex]
[tex]=200 mG-166.67 mG[/tex]
[tex]=33.33 mG[/tex]
[tex]F_{net}=222.33\times 10^{-11} N[/tex]
[tex]F_{net}=2.23m\times 10^{-9} N[/tex]
acceleration [tex]a=\frac{2.23m\times 10^{-9}}{m}[/tex]
[tex]a=2.23\times 10^{-9} m/s^2[/tex]
towards 8 kg mass
(a) The acceleration of the particle is 2.2 x 10⁻⁹ m/s².
(b) The acceleration of the particle is toward the 8.0 kg 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 8.0 kg and mass, m
[tex]F_1 = \frac{6.67 \times 10^{-11} \times 8 \times m}{(0.2)^2} \\\\F_1 = 1.33 \times 10^{-8} m[/tex]
Force between 14 kg and mass, m
distance between 15 kg and m = 50 cm - 20 cm = 30 cm
[tex]F_2 = \frac{6.67 \times 10^{-11} \times 15 \times m}{(0.3)^2} \\\\F_2 = 1.11 \times 10^{-8} m[/tex]
Net force
The net force on the masses;
F(net) = F1 + F2
F(net) = 1.33 x 10⁻⁸m - 1.11 x 10⁻⁸m = 2.2 x 10⁻⁹ m
Acceleration of the particle
F = ma
a = F/m
a = (2.2 x 10⁻⁹m)/m
a = 2.2 x 10⁻⁹ m/s²
Since the acceleration if positive, it will be directed towards 8.0 kg mass.
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