A vertical spring of force constant 100 N/m is attached with a hanging mass of 10 kg. Now…
Physics · JEE Advanced · NTA Exams — Work, Energy and Power
A vertical spring of force constant 100 N/m is attached with a hanging mass of 10 kg. Now an external force is applied on the mass so that the spring is stretched by additional 2 m. The work done by the force F is
(g = 10 m/s2)
(g = 10 m/s2)
- 200 J
- 400 J
- 450 J
- 600 J
Answer
(A) 200 J
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