Question

A small trolley of mass 2 kg resting on a horizontal frictionless turntable is connected by a light spring to the centre of the table. The relaxed length of the spring is 35 cm. when the turntable is rotated an angular frequency of 10 rad $s^{-1}$ the length of the spring becomes 40 cm. what is the force constant of the spring?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The radius of the circle along which the trolley moves is
r = 40cm = 0.4m
When the table is rotated, the tension in the spring is equal to the centripetal force, i.e.
$F=\frac{m{v}^2}{r}=mr{\omega }^2=2\times 0.4\times {10}^3=80N$
The extension in the spring is x = 40−35 = 0.05m
$\therefore$ Force constant $K=\frac{F}{x}=\frac{80}{0.05}=1.6\times {10}^{3}N{m}^{-1}$
Hence the correct choice is (b)