Question

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.

Answer 1

The potential at point A is VA = -5 J/Kg

Correct option is (c).

Explanation:

The work doe y the person is equal to the kinetic energy and potential energy of the mass of 1 kg at point A.

Let VA e the potential at point A.

W =  1/2 mV^2 + (P.E)A

W =  1/2 mV^2 + VA x m

-3 = 1/2  x 1 x (2)^2 + VA x 1

VA = -5 J/Kg

Thus the potential at point A is VA = -5 J/Kg

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Answer 2

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 $-5 J k g-1$.

Explanation:

When the mass moves at the velocity of 2 m s-1, it reaches point A. Please note that work done is equal to the potential energy and kinetic energy of the mass of 1 kg.

$\Delta W=\Delta U+\Delta K$

$\Delta U=\Delta W-\Delta K$

$-3 J=\frac{1}{2} x^{2}(4)+P(A)$

$P(A)=-5 J / k g$

So, the at point A, the potential is -5 J/Kg.