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Two blocks are connected by a massless string that passes over a frictionless peg as shown in figure. One end of the string is attached to a mass m_1=3kg, i.e., a distance R=1.20m from the peg. The other end of the string is connected to a block of mass m_2=6kg resting on a table. From what angle theta, measured from the vertical, must the 3-kg block be released in order to just lift the 6kg block off the table?

Answer» <html><body><p><br/></p>Solution :This problem involves several concepts. First we will apply conservation of energy to find the <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> of the <a href="https://interviewquestions.tuteehub.com/tag/block-18865" style="font-weight:bold;" target="_blank" title="Click to know more about BLOCK">BLOCK</a> `m_1` at the bottom of the circular path as a function of `theta` and the radius of the path, R. <br/> From Newton's second law we will <a href="https://interviewquestions.tuteehub.com/tag/determine-437910" style="font-weight:bold;" target="_blank" title="Click to know more about DETERMINE">DETERMINE</a> the tension at the bottom of its path as function of given <a href="https://interviewquestions.tuteehub.com/tag/parameters-11881" style="font-weight:bold;" target="_blank" title="Click to know more about PARAMETERS">PARAMETERS</a>. Finally, the block `m_2` will lift off the ground when the upward force (tension) exerted by the cord just exceeds the weight of the block. We take bottom of the circle as <a href="https://interviewquestions.tuteehub.com/tag/reference-1181544" style="font-weight:bold;" target="_blank" title="Click to know more about REFERENCE">REFERENCE</a> level. <br/> From conservation of energy, we have <br/> `KE_i+U_i=KE_f+U_f` <br/> `0+m_1g(R-Rcostheta)=1/2m_1v^2+0` <br/> `v^2=2gR(1-costheta)` (i) <br/> Applying Newton's second law on block of mass `m_2`, we have <br/> `sum F_n=T-m_1g=m_1v^2/R` <br/> `T=m_1g+(m_1v^2)/(R)` (ii) <br/> As the string is massless, tenstion T is constant throughout. When `m_2` just lifts off the normal reaction becomes zero. For block `m_2`, we have <br/> `T=m_2g` (iii) <br/> From Eqs. (i), (ii), and (iii), we get <br/> `m_2g=m_1g+m_1(2gR(1-costheta))/(R)` <br/> `cos theta=(3m_1-m_2)/(2m_1)=(3xx3-6)/(2xx3)=1/2` <br/> `theta=60^@`</body></html>


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