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Q.

A small ball of mass M=1kg is attached to two identical springs of constant 100N/m, Which are attached to floor and roof. The springs are initially unstretched. By what distance ball comes down to attain equilibrium? (in cm) ( take g=10ms^-2)

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answer is 5.

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Detailed Solution

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let the ball displaces by $x$ .

The upper spring will elongate by x and lower spring will compress by x.

$Mg - Kx - Kx = 0$

$x = \frac{1 \times 10}{2 \times 100} = 5 cm$

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A small ball of mass M=1kg is attached to two identical springs of constant 100N/m, Which are attached to floor and roof. The springs are initially unstretched. By what distance ball comes down to attain equilibrium? (in cm) ( take g=10ms-2)