00:01
Consider a small block with a mass of 0 .04 kilograms that is moving in the xy plane.
00:08
That force on the block is described by the potential energy function u of xy.
00:16
We want to determine the magnitude and direction of the acceleration of the block when it is at the point x equal to 0 .28 meters and y equal to 0 .57 meters.
00:30
So first we need to recall how the force was so much to a potential energy function is determined.
00:40
So we have that the force is equal to minus the gradient of the potential energy function u.
01:03
So let's evaluate the gradient of our given function.
01:08
This will be equal to the derivative of u with respect to x and the i direction minus the partial derivative of u with respect to y in the j direction.
01:31
So now let's evaluate our partial derivatives.
01:39
So here we have that u of x y is equal to 5 .85 x squared minus 3 .65 y cube.
02:02
So we will have that d 'u over del x will give us 2 times 5 .85 x and we have a del u over the y is equal to minus 3 times 3 .65 y squared.
02:36
So now we can insert this expression into our fourth equation giving us minus 2 times 5 .85 x and the i direction plus three times 3 .65 y squared in the j direction.
03:03
But we specifically want the force at the given coordinates x is equal to 0 .28 meters and y is equal to 0 .57 meters...