"Two blocks of masses m1 and m2 are connected by a spring of spring constant k (figure 9-E15). The block of mass m2 is given a sharp impulse so that it acquires a velocity v0 towards right. Find
(a) the velocity of the center of mass,
(b) the maximum elongation that the spring will suffer."
Consider the situation of the previous problem. Suppose the blocks of mass m1 is pulled by a constant force F1 and the other block is pulled by a constant force F2. Find the maximum elongation that the spring will suffer.
Answers
Thanks for asking the question!
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!
