Respuesta :
Answer:
Acceleration of the particle is 4.45 m/s² and acceleration direction is towards the 8 kg point mass.
Explanation:
Let m₁ be 8 kg point mass and m₂ be 12 kg point mass and take direction along m₁ mass as positive x-axis. The separation between twp point masses is 50 cm. According to the problem, a particle of mass m is held between the two point masses m₁ and m₂.
Distance between m₁ and m, r₁ = 20 cm = 0.2 m
Distance between m₂ and m, r₂ = (50 - 20) cm =30 cm = 0.3 m
The particle of mass m experiences an attractive gravitational force due to both the point masses. Thus, the net force on the particle of mass m is :
F = F₁ + F₂
F₁ is acting along the positive x axis and F₂ is acting along the negative x axis.
[tex]F = \frac{Gm_{1}m }{r_{1} ^{2} } - \frac{Gm_{2}m }{r_{2} ^{2} }[/tex]
Substitute the values of G, m₁, m₂, r₁ and r₂ in the above equation.
[tex]F = \frac{6.67\times10^{-11} \times8m }{0.2 ^{2} } - \frac{6.67\times10^{-11}\times12m }{0.3^{2} }[/tex]
F = 4.45m N/kg
Acceleration of the particle = [tex]\frac{F}{m}[/tex] = 4.45 m/s²
The direction of acceleration of the particle is towards the 8 kg point mass.
