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The breaking strength of the cable used to pull a body is 40 N. A body of mass 8 kg is resting on a table of coefficient of friction mu=0.20. The maximum acceleration which can be produced by the cable is ( take , g=ms^(-2) ) |
Answer» <html><body><p>`6 <a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>^(-2)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> ms^(-2)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> ms^(-2)`<br/>`8 ms^(-2)`</p>Solution :(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `a_("max")=(T_("max")-mu mg)/(m)=()40-0.2xx8xx(10)/(8)=3 ms^(-2)`</body></html> | |