A wire, suspended vertically from one of its ends, is stretched by attaching a weight of $200 N$ to the lower end. The weight stretches the wire by $1 m m$ . The elastic energy (in $J$ ) stored in the wire is
Answer:
0.10
Solution:
$U = \frac{1}{2} \times F \times l = \frac{1}{2} \times 200 \times 10^{- 3} = 0.1 J$

Question

A wire, suspended vertically from one of its ends, is stretched by attaching a weight of $200 N$ to the lower end. The weight stretches the wire by $1 m m$ . The elastic energy (in $J$ ) stored in the wire is

Answer:

0.10

Solution:

$U = \frac{1}{2} \times F \times l = \frac{1}{2} \times 200 \times 10^{- 3} = 0.1 J$

TestHub · Accepted Answer

U = 1 2 × F × l = 1 2 × 200 × 10 - 3 = 0.1 J

Physics - W.E.P. Question with Solution | TestHub

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