A person brings a mass of 1 kg from infinity | vedclass

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(D) According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy: $W_{ext} + W_{grav} = \Delta K.$Here,$W

Vedclass · Accepted Answer

$-10$

Vedclass · Answer

(D) According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy: $W_{ext} + W_{grav} = \Delta K.$Here,$W_{ext} = -5.5 \ J$ is the work done by the person.The work done by the gravitational force is $W_{grav} = -\Delta U = -(U_A - U_{\infty}) = -(m V_A - 0) = -m V_A.$The change in kinetic energy is $\Delta K = K_f - K_i = \frac{1}{2} m v^2 - 0 = \frac{1}{2} 	imes 1 	imes (3)^2 = 4.5 \ J.$Substituting these into the work-energy theorem: $-5.5 + (-1 	imes V_A) = 4.5.$$-V_A = 4.5 + 5.5 = 10 \ J/kg.$Therefore,$V_A = -10 \ J/kg.$

$A$ person brings a mass of $1 \ kg$ from infinity to a point $A.$ Initially,the mass was at rest but it moves at a speed of $3 \ m/s$ as it reaches $A.$ The work done by the person on the mass is $-5.5 \ J.$ The gravitational potential at $A$ is $:$ (in $J/kg$)

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