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 2 m s ⁻¹ as it reaches A. The work done by the person on the mass is −3 J. The potential at A is
(a) −3 J kg⁻¹
(b) −2 J kg⁻¹
(c) −5 J kg⁻⁴
(d) none of these.

Question

Toppr Admin · Accepted Answer

Initially the total energy of the mass is zero-Finally the kinetic energy of mass-12mv2-2-xA0-JWork done on the mass-3-xA0-JTherefore the potential energy of mass at point A-5-xA0-JTherefore the potential at point A is-xA0-5-xA0-J-kg

A person brings a mass of $1 k g$ from infinity to a point A. Initially the mass was at rest but it moves with a speed of $2 m / s$ as it reaches A. The work done by the person on the mass is $3 J .$ The potential of A is.
$7 J / k g$
$3 J / k g$
$2 J / k g$
$5 J / k g$

$Initially the total energy of the mass is zero.$
$Finally the kinetic energy of mass = \frac{1}{2} m v^{2} = 2 J$
$Work done on the mass = 3 J$
$Therefore the potential energy of mass at point A = 5 J$
$Therefore the potential at point A is 5 J/kg$

A person brings a mass of 1kg from infinity to a point A. Initially the mass was at rest but it moves with a speed of 2m/s as it reaches A. The work done by the person on the mass is 3J. The potential of A is.7J/kg3J/kg2J/kg5J/kg

Initially the total energy of the mass is zero.Finally the kinetic energy of mass=12mv2=2 JWork done on the mass=3 JTherefore the potential energy of mass at point A=5 JTherefore the potential at point A is 5 J/kg

A person brings a mass of $1 k g$ from infinity to a point A. Initially the mass was at rest but it moves with a speed of $2 m / s$ as it reaches A. The work done by the person on the mass is $3 J .$ The potential of A is.
$7 J / k g$
$3 J / k g$
$2 J / k g$
$5 J / k g$

$Initially the total energy of the mass is zero.$
$Finally the kinetic energy of mass = \frac{1}{2} m v^{2} = 2 J$
$Work done on the mass = 3 J$
$Therefore the potential energy of mass at point A = 5 J$
$Therefore the potential at point A is 5 J/kg$