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

The diagram showing the variation of gravitational potential of Earth with distance from the centre of Earth is

$A$ particle of mass $m$ is released from infinity and it moves towards a large solid sphere of mass $M$ and radius $R$ made of a jelly-like material. Neglecting the resistance offered by the material of the sphere,find the velocity of the particle when it crosses the centre of the sphere.

The bodies of masses $100 \,kg$ and $8100 \,kg$ are held at a distance of $1 \,m$. The gravitational field at a point on the line joining them is zero. The gravitational potential at that point in $J/kg$ is $\left(G = 6.67 \times 10^{-11} \,Nm^2/kg^2\right)$

On the $x$-axis and at a distance $x$ from the origin,the gravitational field due to a mass distribution is given by $\frac{Ax}{(x^2+a^2)^{3/2}}$ in the $x$-direction. The magnitude of gravitational potential on the $x$-axis at a distance $x$,taking its value to be zero at infinity,is

Give the formula of gravitational potential at distance $r$ $(r > R_E)$ from the centre of the Earth.

Vedclass Test Series

Exam Paper Generator

Online Exam Module