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A body of mass 0.3kg is taken up an inclined plane of length 10m and height 5m, and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the (a) work done by the gravitational force over the round trip. (b) work done by the applied force over the upward journey (c ) work done by frictional force over the round trip. (d) kinetic energy of the body at the end of the trip. How is the answer (d) related to the first three answers ? m=0.3kg, OA=10m, BA=5m |
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Answer» Solution :(a) Gravitational force isconservating force.So work DONE by it for the round trip is ZERO. (b) From FIG.  `OA=10m`, `BA=5m` `OB=sqrt(OA^(2)-BA^(2))` `=sqrt(100-25)` `=sqrt(75)=5sqrt(3)` `costheta=(OB)/(OA)=(5sqrt(3))/(10)=(sqrt(3))/(2)` `sintheta=(BA)/(OA)=(5)/(10)=(1)/(2)` Frictional force `=muN=mu mg costheta` For the upward journey force to be applied `=F=mgsintheta+mu mg costheta` Displacement `s=10m` Work done `=(mgsintheta+mu mg costheta)s` `=mgs(sintheta+mu costheta)` `=0.3xx9.8xx10(0.5+0.15xxsqrt(3)//2)` `=18.52J` (c ) Distance travelled during the round trip `=10+10=20m` Work done by force of friction `=-mu mg costhetaxx20` `=-0.15xx0.3xx9.8xxsqrt(3)//2xx20` `=-7.638J` -ve sign shows that work is done against friction. (d) For the downward motion from A to O.  `u=0`, `v=?` Net acceleration `=a=gsintheta-mu g costheta` `=g(sintheta-mucostheta)` `=9.8(0.5-0.15xxsqrt(3)//2)` `=3.627ms^(-2)` `s=10m` `v^(2)=u^(2)+2as` `=0+2xx3.627xx10` `v^(2)=72.54` K.E.`=1//2mv^(2)` `=1//2xx0.3xx72.54` `=10.88J` From (b) and (c ) Net work done `=18.52-7.64=10.88J` from (d) K.E. `=10.88J` THUS net work done `=` Change in K.E. |
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