Question

Consider the situation of the previous problem. Suppose the block 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.

Answer 1

The maximum elongation that the spring will suffer is explained below

Explanation:

Acceleration of mass , m₁ = $\frac{\left(\mathrm{F}_{1}-\mathrm{F}_{2}\right)}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}$

$\text { Acceleration of mass, } m_{2}=\frac{\left(F_{2}-F_{1}\right)}{\left(m_{1}+m_{2}\right)}$

Mass m1 and m2 block F1 and F2 forces will experience different acceleration and inertia force as well.

Step 2:

Net force on m₁ = F₁ - m₁a = F₂ - m₂  $x \frac{\left(F_{2}-F_{1}\right)}{m_{1}+m_{2}}$

$=\frac{\left(m_{1} F_{2}+m_{2} F_{2}-m_{2} F_{2}+F_{1} m_{2}\right)}{\left(m_{1}+m_{2}\right)}$

$=\frac{\left(m_{1} F_{2}+m_{2} F_{2}\right)}{\left(m_{1}+m_{2}\right)}$

$\text { Net force on } \mathrm{m}_{2}=\mathrm{F}_{2}-\mathrm{m}_{2} \mathrm{a}=\mathrm{F}_{2}-\mathrm{m}_{2} \times \frac{\left(\mathrm{P}_{2}-\mathrm{F}_{1}\right)}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

$=\frac{\left(m_{1} P_{2}+m_{2} F_{2}-m_{2} F_{2}+F_{1} m_{2}\right)}{\left(m_{1}+m_{2}\right)}$

$=\frac{\left(m_{1} F_{2}+m_{2} F_{2}\right)}{\left(m_{1}+m_{2}\right)}$

The mean spring extension is x1 + m2 when the interval between m1 and m2 is x1 and x2 respectively.

Step 3:

Work done by blocks = energy stored in the spring

$=\frac{\left(m_{2} \bar{F}_{1}+m_{1} F_{2}\right)}{m_{1}+m_{2}} \times x_{1}+\frac{\left(m_{2} F_{1}+m_{1} F_{2}\right)}{\left(m_{1}+m_{2}\right)} \times x_{2}=\frac{1}{2} K\left(x_{1}+x_{2}\right)^{2}$

$=\mathrm{x}^{1}+\mathrm{x}^{2}=\frac{2}{\mathrm{K}} \frac{\left(\mathrm{m}_{\mathrm{z}} \mathrm{F}_{1}+\mathrm{m}_{1} \mathrm{p}_{2}\right)}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}$

Answer 2

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

Explanation:

Acceleration of mass m1 = F1 - F2 / m1 + m2

Acceleration of mass m2  = F2 - F1 / m1 + m2

Mass m1 and m2 will experience different acceleration due to F1 and F2 and experience an inertia force.

Net force on m1 = F1 -m1a

= F1 -m1 * F1 - F2 / m1 +  m2

= m1F1 +m2F1 - m1F1 + F2m1 / m1 + m2

= m2F1 + m1F2 / m1 + m2

Net force on m2 = m1F2 + m2F2 / m1 + m2

If displacement by m1 and m2 is x1 and x2 respectively, then the maximum extension of the spring = x1 + m2

Work done by the blocks = energy stored in spring

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

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