Theory question 106
Variation(a) The force of attraction, F, between two bodies varies directly as the product of their masses, m1 and m2 and inversely as the square of the distance, d, between them. Given that F = 20 N, when m1 = 25 kg, m2 = 10 kg and d = 5 m, find: (i) an expression for F in terms of m1, m2 and d; (ii) the distance, d when F = 30 N, m1 = 7.5 kg and m2 = 4 kg.
(b) Find the value of x in the diagram.
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Model answer
(a)(i) F = km1m2/d2; 20 = k(25)(10)/25 → k = 2
F = 2m1m2/d2
(ii) 30 = 2(7.5)(4)/d2 = 60/d2 → d2 = 2
d = √2 m ≈ 1.41 m
(b) The figure is a pentagon: angle sum = (5 − 2) × 180° = 540°
x + (x + 20) + (x + 40) + (x + 60) + (x + 80) = 540 → 5x + 200 = 540
x = 68°
Explanation
Find the constant k first, then use the formula with the new values.