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Question: The gravitational field in a region is given by the equation \(\vec { E } =\left( 5\hat { i } +12\ha...

The gravitational field in a region is given by the equation E⃗=(5i^+12j^)\vec { E } =\left( 5\hat { i } +12\hat { j } \right) N/kg. If the particle of mass 2kg is moved from the origin to the point (12m, 5m) in this region, the change in gravitational potential energy is
A. -225J
B. -240J
C. -245J
D. -250J

Explanation

Solution

Hint: Gravitational potential energy is the energy that an object has due to its position or place in the gravitational force area. To find the difference in potential energy, we should use general expression derived from the gravitational law.
Formula used: Ex=−dvdr{ E }_{ x }=-\dfrac { dv }{ dr }

Complete step by step answer:
The given equation of gravitational field is
E⃗=(5i^+12j^)\vec { E } =\left( 5\hat { i } +12\hat { j } \right) It means that the potential is in two directions: x and y
First, let us find the potential in x – direction:
We know, Ex=−dvdr{ E }_{ x }=-\dfrac { dv }{ dr }
It can also be written as, dv = -E.dr
On integrating both sides,
∫dv=−∫Exdx\int { dv } =-\int { { E }_{ x } } dx
Substituting the value of Ex=5{ E }_{ x }=5
∫dv=−∫5dx\int { dv } =-\int { 5 } dx
Vx=−5x{ V }_{ x }=-5x, now substitute x = 12 in the equation
Vx=−60V{ V }_{ x }=-60V.
Now, finding the potential in y – direction:
As seen before, Ey=−dvdr{ E }_{ y }=-\dfrac { dv }{ dr }
It can also be written as, dv = -E.dr
On integrating both sides,
∫dv=−Eydy\int{dv}=-{{E}_{y}}dy
Substituting the value of Ey=12{ E }_{ y }=12
∫dv=−∫12dy\int{dv}=-\int{12}dy
Vy=−12y{{V}_{y}}=-12y, now substitute y = 5 in the equation
Vy=−60V{ V }_{ y }=-60V.
The total potential energy is given by V=Vx+Vy{ V }={ V }_{ x }+{ V }_{ y }
V = -60-60 = -120V
The change in gravitational potential energy, work done = m×ΔV m\times \Delta V
The given mass of the particle, m = 2kg
So, work done = 2 × -120 = -240 J
Therefore, the correct answer for the given question is option (B).

Note: Gravitational potential energy is often used for the objects that are on the earth’s surface as it can easily be considered as 9.8m/s2. The potential energy above a particular height is equal to the work done to raise the object to that height without losing any form of kinetic energy.