question_answer 74)

                  A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of \[{{30}^{o}}\] by a force of 10 N parallel to the inclined surface (Fig.).The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate                 (a) work done against gravity                 (b) work done against force of friction                 (c) increase in potential energy                 (d) increase in kinetic energy                 (e) work done by applied force.                

Answer:

                      (a) Work done against gravity = Force of gravity \[\times S\]                 \[=mg\text{ }\sin \,\theta \times S\]                 \[=1\times 10\times \sin \text{ }30{}^\circ \times 10\]                 \[=10\times \frac{1}{2}\,\times \,10\,=50\,J\] (b)          Force of friction \[f=\mu R\]             \[=\mu \text{ }Mg\text{ }\cos 30{}^\circ \]                 \ Work done against force of friction                 \[=f\times S\]                 \[=\mu \,mg\,\cos {{30}^{o}}\times 5\]                 \[=0.1\times 1\times 10\times \frac{\sqrt{3}}{2}\times 10=8.66\text{ }J\] (c)           Height through which block is raised,                 \[h=10\text{ }\sin 30{}^\circ =5\text{ m}\]                 \ Increase in P.E. \[=mgh=1\times 10\times 5\]                 =  50 J (d)          Total force acting on the block                 \[=Fmg\text{ }\sin \text{ }30{}^\circ -f\]                 \[=Fmg\text{ }\sin \text{ }30{}^\circ -\mu R\]                 \[=Fmg\text{ }\sin \text{ }30{}^\circ -\mu \,\,mg\,\,\cos {{30}^{o}}\]             \[=Fmg\text{ (}\sin 30{}^\circ +\mu \,\,\cos {{30}^{o}})\]             \[=10-1\times 10\left( \frac{1}{2}+0.1\times \frac{\sqrt{3}}{2} \right)\]             or \[F=4.1\,N\]                 Increase in K.E. = Work done \[=F'\times S\]                 \[=4.1\times 10=41J\]                 (e) Work done by applied force \[=F\times S\]                 \[=10\times 10=100\text{ }J\]