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
[tex]A=\sqrt{a^2+a'^2}[/tex]
[tex]A=\sqrt{120^2+88.2^2}[/tex]
A= 148.92 m/s²
The magnitude of the acceleration at that point is:
|a| = 14.3 J/m*kg
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:
[tex]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)[/tex]
Then at the point x = 0.40m, y = 0.56m the force is:
[tex]F = (-4.5 J/m, 3.528 J/m)[/tex]
Then the acceleration at that point is:
[tex]a = F/0.04kg = (-4.5 J/m, 3.528 J/m)*\frac{1}{0.04kg}[/tex]
The magnitude of the acceleration is:
[tex]|a| = \sqrt{(\frac{1}{0.4kg})^2*( (-4.5 J/m)^2 + (3.528 J/m)^2) } = 14.3 J/m*kg[/tex]
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