Solveeit Logo

Question

Question: A person brings a mass of \(1kg\) from infinity to a point \(A\). Initially the mass was at rest but...

A person brings a mass of 1kg1kg from infinity to a point AA. Initially the mass was at rest but it moves at a speed of 2ms12m{s^{ - 1}} as it reaches AA. The work done by a person on the mass is 3J3J. The potential at AA is:
(A) 3J/kg - 3J/kg
(B) 2J/kg - 2J/kg
(C) 5J/kg - 5J/kg
(D) None of these

Explanation

Solution

We are given with the mass of the test body, its velocity and the work done by the test body and are asked about the potential energy per unit mass of the body. Thus, we will apply the formula for finding the total work done by the body.
Formulae Used:
W=T+UW = T + U
Where, WW is the work done by the body,TT is the kinetic energy of the body and UU is the potential energy of the body.
U(m)=UmU\left( m \right) = \dfrac{U}{m}
Where,U(m)U\left( m \right) is the potential energy per unit mass andmm is the mass of the body.
T=12mv2T = \dfrac{1}{2}m{v^2}
Where,vv is the velocity of the body.

Complete step by step answer
Here,
Work done,W=3JW = - 3J
Velocity of the body,v=2ms1v = 2m{s^{ - 1}}
Mass of the body,m=1kgm = 1kg
Now,
T=12mv2T = \dfrac{1}{2}m{v^2}
Further,
T=12(1)(2)2T = \dfrac{1}{2}\left( 1 \right){\left( 2 \right)^2}
Then, we get
T=2JT = 2J
Now,
As per the formula,
W=T+UW = T + U
Then, we get
U=WTU = W - T
Now,
Substituting the values, we get
U=32U = - 3 - 2
Further, we get
U=5JU = - 5J
Now,
For the potential energy per unit mass, we will use
U(m)=UmU\left( m \right) = \dfrac{U}{m}
Then,
Substituting the values, we get
U(m)=51U\left( m \right) = \dfrac{{ - 5}}{1}
Thus,
U(m)=5J/kgU\left( m \right) = - 5J/kg

Hence, the correct answer is (C).

Note We have taken the value of work as we are equating it with the total energy of the body and as per definition, the total energy of a body is the negative of the work done by it. Thus, we take the negative of the value of the work.