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padilas [110]
3 years ago
15

An 8.00 kg point mass and a 15.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

tween the two masses 20.0 cm from the 8.00 kg mass along the line connecting the two fixed masses.
Find the magnitude and direction of the acceleration of the particle. Answer from book: 2.2 x 10^-9 m/s^2 - toward the 8.00 kg mass....
Physics
1 answer:
Verdich [7]3 years ago
3 0

Answer:a=2.23\times 10^{-9} m/s^2

Explanation:

Given

Mass of first Point mass m_1=8 kg

Mass of second Point m_2=15 kg

distance between them d=50 cm

third point mass m_3=m

Distance between m\ and\ m_1 is\ 20 cm

Distance between  m\ and\ m_2 is 30 cm

Force Due to m_1\ and\ m F_1=\frac{Gmm_1}{d_1^2}

F_1=\frac{8Gm}{(0.2)^2}

F_1=200 mG

F_2=m\frac{Gmm_2}{d_2^2}

F_1=m\frac{15Gm}{(0.3)^2}

F_2=166.67 mG

Net Force

F_{net}=F_1-F_2

=200 mG-166.67 mG

=33.33 mG

F_{net}=222.33\times 10^{-11} N

F_{net}=2.23m\times 10^{-9} N

acceleration a=\frac{2.23m\times 10^{-9}}{m}

a=2.23\times 10^{-9} m/s^2

towards 8 kg mass

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