The potential energy function of a particle due to some gravitational field is given by `U=6x+4y`. The mass of the particle is `1kg` and no other force is acting on the particle. The particle was initially at rest at a point `(6,8)`. Then the time (in sec). after which this particle is going to cross the `'x'` axis would be:

Answer

2

Explanation

The force acting on the particle can be given by
`vec(F)=(-delU)/(delr)=-6hat(i)-4hat(j)`
Thus acceleration of this particle would be
`a_(x)=6m^(-2)` and `a_(y)=4ms^(-2)`
when the particle crosses the `'x'` axis
`8=1/2xx4(t^(2))impliest^(2)=4`
`t=2s`