The length of a steel wire increases by $0.5 c m$ when it is loaded with a weight of $9 k g$ . Calculate the force constant of the wire and W.D. in stretching wire.

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Solution

The force constant is given as,

$k = \frac{m g}{x}$

$k = \frac{9 \times 10}{0.5 \times 10^{- 2}}$

$k = 18 \times 10^{3} N / m$

The work done is given as,

$W = \frac{1}{2} k x^{2}$

$W = \frac{1}{2} \times 18 \times 10^{3} {(0.5 \times 10^{- 2})}^{2}$

$W = 0.225 J$

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The length of a steel wire increases by 0.5cm when it is loaded with a weight of 9kg . Calculate the force constant of the wire and W.D. in stretching wire.

The force constant is given as, k=mgx k=9×100.5×10−2 k=18×103N/m The work done is given as, W=12kx2 W=12×18×103(0.5×10−2)2 W=0.225J

The length of a steel wire increases by $0.5 c m$ when it is loaded with a weight of $9 k g$ . Calculate the force constant of the wire and W.D. in stretching wire.

The force constant is given as,

$k = \frac{m g}{x}$

$k = \frac{9 \times 10}{0.5 \times 10^{- 2}}$

$k = 18 \times 10^{3} N / m$

The work done is given as,

$W = \frac{1}{2} k x^{2}$

$W = \frac{1}{2} \times 18 \times 10^{3} {(0.5 \times 10^{- 2})}^{2}$

$W = 0.225 J$