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A block of mass 10kg is dragged across a horizontal surface by pulling the blockwith force of 100N. The force is applied by attaching a chord to the block. The chord is inclined at an angle of 60^(@) with the horizontal. If the coefficient of kinetic friction is 0.2 what is theacceleration of the block ?

Answer» <html><body><p></p>Solution :`m=10kg`, `F=100N` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a>=60^(@)`, `mu_(<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>)=0.2`, `a=?` <br/> Horizontal component of the applied force <br/> `=F cos theta` <br/> `=100xxcos60=50N` <br/> <a href="https://interviewquestions.tuteehub.com/tag/frictional-465947" style="font-weight:bold;" target="_blank" title="Click to know more about FRICTIONAL">FRICTIONAL</a> force opposing the motion of the block <br/> `=mu_(k)mg` <br/> `=0.2xx10xx9.8=19.6N` <br/> <a href="https://interviewquestions.tuteehub.com/tag/net-5194" style="font-weight:bold;" target="_blank" title="Click to know more about NET">NET</a> force acting on the block `=Fcostheta-mu_(k)mg` <br/> `=50-19.6=30.4N` <br/> <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> of the block `=("Net force")/("Mass")=(30.4)/(10)=3.04m//s^(2)` <br/> `=3.04m//s^(2)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/SCH_DKB_ISC_PHY_XI_C05_SLV_003_S01.png" width="80%"/></body></html>


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