Question

What is the magnitude of the acceleration of the block when it is at the point `x= 0.24 m`, `y= 0.52m`? What is the direction of the acceleration of the block when it is at the point `x= 0.24m`, `y= 0.52m`? (See details).

A small block with mass
`0.0400 kg` is moving in the xy-plane. The net force on the block is described by the potential- energy function `U(x,y)= (5.90 J/m^2 )x^2-(3.65 J/m^3 )y^3`.

Answer 1

Since `xand y` are orthogonal to each other these can be treated independently. We also know that

`vecF=-gradU`

`:.x`-component of two dimensional force is

`F_x = -(delU)/(delx)`
`F_x = -del/(delx)[(5.90\ Jm^-2)x^2−(3.65\ Jm^-3)y^3]`
`F_x = -11.80x`

`x`-component of acceleration

`F_x=ma_x = -11.80x`
`0.0400a_x = -11.80x`
`=>a_x = -11.80/0.0400x`
`=>a_x = -295x`

At the desired point

`a_x = -295xx0.24`
`a_x = -70.8\ ms^-2`

Similarly `y`-component of force is

`F_y = -del/(dely)[(5.90\ Jm^-2)x^2−(3.65\ Jm^-3)y^3]`
`F_y = 10.95y^2`

`y`-component of acceleration

`F_y=ma_ = 10.95y^2`
`0.0400a_y = 10.95y^2`
`=>a_y = 10.95/0.0400y^2`
`=>a_y =27.375y^2`

At the desired point

`a_y = 27.375xx(0.52)^2`
`a_y = 7.4022\ ms^-2`

Now `|veca| = sqrt[a_x^2 + a_y^2]`

`|veca| = sqrt[(-70.8)^2 + (7.4022)^2]`
`|veca| =71.2\ ms^-2`

If `theta` is the angle made by acceleration with `x`-axis at the desired point then

`tantheta = (a_y)/(a_x)`

Inserting calculated values

`tantheta = (7.4022)/(-70.8)`, (`2nd` quadrant)
`=>theta=174^@`