Question

A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy function U(x,y) = (6.00 J/m2 )x 2 - (3.75 J/m3 )y 3 . What is the magnitude of the acceleration (in m/s2 ) of the block when it is at the point x = 0.40 m, y = 0.56 m?

Answer 1

Answer:

A= 148.92  m/s²

Explanation:

Given that

U(x,y) = (6.00  )x²  - (3.75  )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

- 4.8 = 0.04 x a

a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

$A=\sqrt{a^2+a'^2}$

$A=\sqrt{120^2+88.2^2}$

A= 148.92  m/s²