First slide

Work done by Diff type of forces

Question

The spring extends by x on loading, then energy stored by the spring is If (If T is the tension in spring and ft is spring constant)

Moderate

A

$T^{2} / 2 k$
B

$T^{2} / 2 k^{2}$
C

$2 k / T^{2}$
D

$2 T^{2} / k$

Solution

Energy stored in the spring,
$U = \frac{1}{2} {kx}^{2} where k is spring constant According to Hooke' s law, T = k \times ∴ x = (T / k) From eqs . (1) and (2), we 8 et U = \frac{1}{2} (\frac{T^{2}}{k})$

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