Question

Two blocks of masses m1 and m2 connected with a light spring of spring constant k are acted by forces F1 and F2 on a frictionless horizontal surface. Find the spring force at this instant provided that at this instant the Acceleration of both the blocks are same.​

Answer 1

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Answer 2

Answer:

Explanation:

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ANSWER::

Acceleration of mass , m₁ = (F₁-F₂) / (m₁ + m₂)

Acceleration of mass , m₂ = (F₂-F₁) / (m₁ + m₂)

Due to forces F₁ and F₂ block of mass m₁ and m₂ will experience different acceleration and will also experience an inertia force.

Net force on m₁ = F₁ - m₁a = F₂ - m₂ x [ (F₂-F₁) / m₁ + m₂]

= (m₁F₂ + m₂F₂ - m₂F₂ + F₁m₂) / (m₁ + m₂)

= (m₁F₂ + m₂F₂) / (m₁+m₂)

Net force on m₂ = F₂ - m₂a = F₂ - m₂ x [ (F₂-F₁) / m₁ + m₂]

= (m₁F₂ + m₂F₂ - m₂F₂ + F₁m₂) / (m₁ + m₂)

= (m₁F₂ + m₂F₂) / (m₁+m₂)

If m₁ and m₂ displaces by a distance of x₁ and x₂ respectively then the maximum extension of spring is x₁ + m₂

Work done by blocks = Energy stored in spring

[(m₂F₁ + m₁F₂) / (m₁ + m₂)] x x₁ + [(m₂F₁ + m₁F₂) / (m₁ + m₂)] x x₂ = (1/2) K (x₁ + x₂)²

x₁ + x₂ = (2/K)[(m₂F₁ + m₁F₂) / (m₁ + m₂)]

Hope it helps!