Physics, asked by skzaid843, 10 months ago

Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force F instead of any impulse. Find the maximum elongation that the spring will suffer and the distance moved by the two blocks in the process.

Answers

Answered by shilpa85475
2

Explanation:

As both blocks are pulled by the same force, they move suddenly with some acceleration and stop instantly at the same position where maximum spring elongation is required.

Let block m1 and m2 expansion be x1 and x2

Total work done = Fx₁ + Fx₂ …………………….eq^n(1)

Increase Potential Energy of spring= \frac{1}{2} K\left(x_{1}+x_{2}\right)^{2} \ldots \ldots \ldots \text { eq }^{n}(2)

Step 2:

Equating Equation 1 and Equation 2

F\left(x_{1}+x_{2}\right)=\frac{1}{2} K\left(x_{1}+x_{2}\right)^{2}

\left(x_{1}+x_{2}\right)=\frac{2 F}{K}

Step 3:

Since net external force is zero on two points, the same force operates in the opposite direction.  

Then,

m_{1} x_{1}=m_{2} x_{2} \ldots \ldots \ldots \ldots \ldots \ldots \text { eq }^{n}(3)

\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)=\frac{2 \mathrm{F}}{\mathrm{K}} \text { (as shown above) }

\mathrm{X}_{2}=\frac{\mathrm{m}_{1} \mathrm{x}_{2}}{\mathrm{m}_{2}}

Step 4:

Substituting , \left(\frac{m_{1} x_{1}}{m_{2}}\right)+x_{1}=\frac{2 F}{K}

\begin{aligned}&x_{1}\left(1+\frac{m_{1}}{m_{2}}\right)=\frac{2 F}{K}\\&x_{1}=\frac{2 F m_{2}}{K\left(m_{1}+m_{2}\right)}\end{aligned}

Similarly ,

x_{2}=\frac{2 \mathrm{F} \mathrm{m}_{1}}{\mathrm{K}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}

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