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JEE · NEET Previous Year Question

$The work done in stretching a spring of natural length 25 cm and spring constant 50 Nm^{- 1} from 50 cm to 60 cm is$

Topic: JEE PYQ

Answer: A. Full step-by-step solution below.

A $1.5 J$
B $2 J$
C $3.5 J$
D $5 J$

Correct answer: A

Solution

(a) Δ x_{i} = (50 - 25) cm = 25 cm Δ x_{f} = (60 - 25) cm = 35 c m So, work done = Δ U = \frac{1}{2} K (Δ x_{f}^{2} - Δ x_{i}^{2}) = \frac{1}{2} \times 50 (35^{2} - 25^{2}) \times 10^{- 4} = 1.5 J

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