Q29 / 92 · NTA Abhyas (JEE Main) 2020 · Physics · Work Power Energy

A vertical spring of force constant 100 Nm-1 is attached to the ceiling, with a block of mass 10 kg suspended from it. Now an external force F is applied on the block so that the spring is stretched by an additional length of 2 m. The work done by the force F is [Take g=10 ms-2]


Figure

  • Correct answer
    A
    200 J
  • B
    400 J
  • C
    100 J
  • D
    600 J

Solution

x = elongation in spring due to mass 10 kg

= 1 0 × 1 0 1 0 0 = 1 m

W F = 1 2 × 1 0 0 × 3 2 - 1 2 - 1 0 × 1 0 × 2

= 200 J

Hence, the correct answer is option: 'A'.

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