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

Answer 2

The magnitude of the acceleration at that point is:

|a| = 14.3 J/m*kg

How to get the acceleration?

Remember that the relation between acceleration and force is given by:

F = a*m

Where F is force, a is acceleration and m is mass.

Solving for the acceleration, we get:

a = F/m.

In this case, we have:

m = 0.04 kg

And the potential energy.

U = (6 J/m^2)*x^2 - (3.75 J/m^3)*y^3

The force is minus the gradient of the potential energy:

$F = (-\frac{dU}{dx} , -\frac{dU}{dy} )\\\\F = ( -2*6.00 J/m^2*x, 3*3.75 J/m^3*y^2)\\\\F = (-12.00 J/m^2*x, 11.25 J/m^3*y^2)$

Then at the point x = 0.40m, y = 0.56m the force is:

$F = (-4.5 J/m, 3.528 J/m)$

Then the acceleration at that point is:

$a = F/0.04kg = (-4.5 J/m, 3.528 J/m)*\frac{1}{0.04kg}$

The magnitude of the acceleration is:

$|a| = \sqrt{(\frac{1}{0.4kg})^2*( (-4.5 J/m)^2 + (3.528 J/m)^2) } = 14.3 J/m*kg$

If you want to learn more about acceleration, you can read:

https://brainly.com/question/19365526