Test SeriesFree PYQsExamsPricing

JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

A spring of spring constant 5 × 10³ N m⁻¹ is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

Answer: (B)

Solution

Work done in stretching from x₁ to x₂ is W = ½k(x₂² − x₁²). With x₁ =

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-153535