Physics, asked by kamran8532, 10 months ago

The block of mass m1 shown in figure (12−E2) is fastened to the spring and the block of mass m2 is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k) (m1 + m2)g sin θ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?
Figure

Answers

Answered by shilpa85475
1

Explanation:

(a) In equilibrium condition, k * x=\left(m_{1}+m_{2}\right) g \sin \theta.  Thus, the compression of the spring is x=\frac{\left(m_{1}+m_{2}\right) g \sin \theta}{k} .

(b) We know that \omega=\sqrt{\frac{k}{m_{1}+m_{2}}} \text { and } x_{1}=\frac{2+\left(m_{1}+m_{2}\right) \varepsilon \sin \theta}{k}.  When the blocks are pushed further distance against the spring and released, they become separated at P=0. So, \mathrm{m}_{2} \mathrm{g} \sin \theta=\mathrm{m}_{2} \mathrm{x}_{2} \omega^{2}.

Therefore, \mathrm{x}_{2}=\frac{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{z} \sin \theta}{\mathrm{k}}. This shows that the blocks will get separated when the springs attain their natural length.

(c) At the time of separation, the total compression is  \mathrm{x}_{1}+\mathrm{x}_{2}. Therefore,  \frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}-0=\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}-\left(m_{1}+m_{2}\right) g \sin \theta\left(x_{1}+x_{2}\right) where v is the velocity of the blocks. Thus, on solving, we get the common speed of the blocks v=\sqrt{\frac{3}{k}\left(m_{1}+m_{2}\right)} g \sin \theta.

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