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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $J/kg$)

Similar Questions

In a gravitational field,at a point where the gravitational potential is zero,

What is the binding energy of a body of mass $m$ lying on the surface of the Earth at a distance $r$ from the center of the Earth?

The $S.I.$ unit of gravitational potential is

If the gravitational field in space is given as $E = -\frac{K}{r^2}$. Taking the reference point at $r = 2\,cm$ with gravitational potential $V = 10\,J/kg$. Find the gravitational potential at $r = 3\,cm$ in $SI$ units. (Given: $K = 6\,J\cdot cm/kg$)

$A$ body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth $(R_e)$. The increase in potential energy will be . . . . . . . ($g$ is acceleration due to gravity at the surface of earth)

Vedclass Test Series

Exam Paper Generator

Online Exam Module