The force of attraction F, between two bodies, varies directly as the product of their masses, \(m_1\) and m\(_2\) and inversely as the square of the distan...
Question 1 Report
The force of attraction F, between two bodies, varies directly as the product of their masses, \(m_1\) and m\(_2\) and inversely as the square of the distance, d, between them. Given that F = 20N, when m\(_1\) = 25kg, m\(_2\) = 10kg and d = 5m, find:
(I) an expression for F in terms of m\(_1\), m\(_2\) and d;
(ii) the distance, d for F = 30N, m\(_1\) = 7.5kg and m\(_2\) = 4kg
(b) Find the value of x in the diagram
Answer Details
(a) Setting up the joint variation
F varies directly as \(m_1 m_2\) and inversely as \(d^2\):
\[F = \frac{k\, m_1 m_2}{d^2}\]
(i) Finding k and the expression
Substitute \(F = 20,\ m_1 = 25,\ m_2 = 10,\ d = 5\):
\[20 = \frac{k(25)(10)}{5^2} = \frac{250k}{25} = 10k\]
\[k = 2\]
Therefore the required expression is
\[F = \frac{2\,m_1 m_2}{d^2}\]
(ii) Finding d
Substitute \(F = 30,\ m_1 = 7.5,\ m_2 = 4\):
\[30 = \frac{2(7.5)(4)}{d^2} = \frac{60}{d^2}\]
\[d^2 = \frac{60}{30} = 2\]
\[d = \sqrt{2} \approx 1.41\ \text{m}\]
(b) Finding x in the diagram
The figure is a five-sided polygon (pentagon) whose interior angles, read from the diagram, are:
\[x,\quad (x+20^\circ),\quad (x+40^\circ),\quad (x+80^\circ),\quad (x+60^\circ)\]
The sum of the interior angles of a pentagon is
\[(5-2)\times 180^\circ = 540^\circ\]
So:
\[x + (x+20^\circ) + (x+40^\circ) + (x+80^\circ) + (x+60^\circ) = 540^\circ\]
\[5x + 200^\circ = 540^\circ\]
\[5x = 340^\circ\]
\[x = 68^\circ\]
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