Three masses m, 2m and 3m are arranged in two | vedclass

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

Question

(D) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W = U_f - U_i$.For configuration $

Vedclass · Accepted Answer

$-\frac{G m^2}{a}\left[6-\frac{6}{\sqrt{2}}\right]$

Vedclass · Answer

(D) The work done by an external agent is equal to the change in the gravitational potential energy of the system: $W = U_f - U_i$.For configuration $1$ (right-angled triangle with sides $a, a, \sqrt{2}a$):$U_i = -\frac{G(m)(2m)}{a} - \frac{G(m)(3m)}{a} - \frac{G(2m)(3m)}{\sqrt{2}a} = -\frac{G m^2}{a} \left( 2 + 3 + \frac{6}{\sqrt{2}} \right) = -\frac{G m^2}{a} \left( 5 + \frac{6}{\sqrt{2}} \right)$.For configuration $2$ (equilateral triangle with all sides $a$):$U_f = -\frac{G(m)(3m)}{a} - \frac{G(m)(2m)}{a} - \frac{G(2m)(3m)}{a} = -\frac{G m^2}{a} (3 + 2 + 6) = -\frac{11 G m^2}{a}$.Work done $W = U_f - U_i = -\frac{11 G m^2}{a} - \left( -\frac{G m^2}{a} \left( 5 + \frac{6}{\sqrt{2}} \right) \right) = \frac{G m^2}{a} \left( -11 + 5 + \frac{6}{\sqrt{2}} \right) = \frac{G m^2}{a} \left( \frac{6}{\sqrt{2}} - 6 \right) = -\frac{G m^2}{a} \left( 6 - \frac{6}{\sqrt{2}} \right)$.

Three masses $m, 2m$ and $3m$ are arranged in two triangular configurations as shown in figure $1$ and figure $2$. The work done by an external agent in changing the configuration from figure $1$ to figure $2$ is:

Similar Questions

An object is taken to a height of $2R$ above the surface of the Earth. What is the increase in its gravitational potential energy? [$R$ is the radius of the Earth]

The gravitational field in a region is given by the equation $E = (5\hat{i} + 12\hat{j}) \,N/kg$. If a particle of mass $2 \,kg$ is moved from the origin to the point $(12 \,m, 5 \,m)$ in this region, what is the change in gravitational potential energy (in $\,J$)?

Two particles of mass $m$ and $9m$ are separated by a distance $r$. At a point on the line joining them,the gravitational field is zero. What is the gravitational potential at that point? ($G$ is the universal gravitational constant)

$A$ 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