A porter lifts a heavy suitcase of mass 80, t | vedclass

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(D) According to the Work-Energy Theorem,the net work done on an object is equal to the change in its kinetic energy.Since the suitcase is moved with

Vedclass · Accepted Answer

$-627.2\, 	ext{J}$

Vedclass · Answer

(D) According to the Work-Energy Theorem,the net work done on an object is equal to the change in its kinetic energy.Since the suitcase is moved with a constant velocity,the change in kinetic energy $\Delta K.E. = 0$.Therefore,the total work done by all forces is $W_{	ext{Porter}} + W_{	ext{gravity}} = 0$.This implies $W_{	ext{Porter}} = -W_{	ext{gravity}}$.The work done by gravity is $W_{	ext{gravity}} = mgh$,where $h$ is the displacement.Here,$m = 80\, 	ext{kg}$,$g = 9.8\, 	ext{m/s}^2$,and $h = 80\, 	ext{cm} = 0.8\, 	ext{m}$.Since the porter is lowering the suitcase,the force applied by the porter is upwards while the displacement is downwards.$W_{	ext{Porter}} = -mgh = -(80) 	imes (9.8) 	imes (0.8) = -627.2\, 	ext{J}$.

$A$ porter lifts a heavy suitcase of mass $80\, \text{kg}$ and at the destination lowers it down by a distance of $80\, \text{cm}$ with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (Take $g = 9.8\, \text{m/s}^2$) (In $\text{J}$)

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