Find the work done by friction if a 1 , kg bl | vedclass

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(A) According to the Work-Energy Theorem,the net work done on the block is equal to the change in its kinetic energy.Since the block moves with a cons

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$-50 \, J$

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(A) According to the Work-Energy Theorem,the net work done on the block is equal to the change in its kinetic energy.Since the block moves with a constant velocity,the change in kinetic energy $\Delta KE = 0$.Therefore,the net work done $W_{net} = W_{gravity} + W_{friction} = 0$.$W_{friction} = -W_{gravity}$.The work done by gravity is $W_{gravity} = (mg \sin 	heta) 	imes d$,where $m = 1 \, kg$,$g = 10 \, m/s^2$,$	heta = 30^{\circ}$,and $d = 10 \, m$.$W_{gravity} = 1 	imes 10 	imes \sin(30^{\circ}) 	imes 10 = 10 	imes 0.5 	imes 10 = 50 \, J$.Thus,$W_{friction} = -50 \, J$.

Find the work done by friction if a $1 \, kg$ block reaches the end of an inclined plane of length $10 \, m$ and inclination $30^{\circ}$ with constant velocity.

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