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icang [17]
3 years ago
5

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

ese objects on a 61.0-kg object placed midway between them. 1 .0000003383 Correct: Your answer is correct. N (b) At what position (other than an infinitely remote one) can the 61.0-kg object be placed so as to experience a net force of zero from the other two objects? 2 1.824 Incorrect: Your answer is incorrect.
Physics
1 answer:
pav-90 [236]3 years ago
4 0

Answer:F_{net}=3.383\times 10^{-7}\ N

Explanation:

Given

Mass of first object m_1=225\ kg

Mass of second object m_2=525\ kg

Distance between them d=3.8\ m

m_3=61\ kg object is placed between them

So force exerted by m_1 on m_3

F_{13}=\frac{Gm_1m_3}{1.9^2}

F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}

F_{13}=2.5374141274×10^{−7}\ N

Force exerted by m_2\ on\ m_3

F_{23}=\frac{Gm_2m_3}{1.9^2}

F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}

F_{23}=5.920632964\times 10^{-7}\ N

So net force on m_3 is

F_{net}=F_{23}-F_{13}

F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}

F_{net}=3.383\times 10^{-7}\ N

i.e. net force is towards m_2

(b)For net force to be zero on m_3, suppose

So force exerted by m_1 and m_2 must be equal

F_{13}=F_{23}

\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}

\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}

\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}

\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}

\Rightarrow 3.8-x=1.52752x

\Rightarrow 3.8=2.52x

\Rightarrow x=1.507\ m

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Answer:v=\sqrt{\dfrac{2[mgh+fh]}{m}}

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