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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

A particle of mass 2×10⁻² kg is moving along the positive X-axis under the influence of a force F(x) = −k/(4x²), where k = 10⁻² Nm². At time t=0, it is at x=1.5 m and its velocity v=0. Find its speed (in m/s) when it reaches x=0.25 m. (Take √30 = 5.5)

Answer: 0.92

Solution

By the work-energy theorem, ½mv² = ∫F dx from x=1.5 to x=0.25 = −(k/4)∫(1/x²)dx from 1.5 to 0.25 = (k/4)[1/x] from 1.5 to 0.25 = (k/4)(1/0.25 − 1/1.5) = (k/4)(4 −

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Question ID VK-154235