Question

Question #e4198

Answer 1

We know that the conservative force `vecF` is related to the potential energy function U(x,y) in two dimension through the following relationship

`vecF=-(delU)/(delx)hati-(delU)/(dely)hatj`

Here it is given that `U=6x+6y`

So

`vecF=-(del(6x+8y))/(delx)hati-(del(6x+8y))/(dely)hatj`

`=>vecF=(-6hati-8hatj)N`

So acceleration `veca=(vecF)/M`

where `M=` mass of the body moving `=2kg`

Hence `veca=(vecF)/2=(-3hati-4hatj)m"/"s^2`

When the body crosses the Y-axis, then x=0.
Again it is given that body is at rest at the position (6m,4m) at t=0
This means
at x=6m, velocity =0;
When the body crosses Y-axis,its displacement along X-axis becomes `s_x=-6m`.

If this displacement occurs in t sec due to its movement with acceleration along X axis `a_x=-3m"/"s^2`

then we can write

`s_x=1/2a_xt^2`

`=>-6=1/2xx(-3)t^2`

`=>t=2s`