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A small object with mass m = 0.0900 kg moves in the xy-plane. The only force on the object is a conservative force that has the potential-energy function U(x,y) = -αx² + βy³, where

α = 2.00 J/m² and β = 0.300 J/m³. The object is released from rest at x = 0.0 m and y = 0.0 m.

a. What is the speed of the object when it is at x = 4.00 m, and y = 0.30 m?

b. What is the acceleration (in unit vectors) of the object when it is at x = 4.00 m, and y = 0.30 m?

User Bouvierr
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Final answer:

The speed of the object when it is at x = 4.00 m and y = 0.30 m is approximately 33.51 m/s. The acceleration of the object at this position is approximately -177.78î + 0.09ĵ m/s².

Step-by-step explanation:

In order to find the speed of the object at x = 4.00 m and y = 0.30 m, we need to calculate the total mechanical energy of the object at this position and then use the conservation of energy principle. The formula for potential energy in this case is U(x,y) = -αx² + βy³. Plugging in the given values, we get U(4.00, 0.30) = -2.00(4.00)² + 0.300(0.30)³ = -32.00 + 0.0081 = -31.99 J.

Next, we need to find the kinetic energy of the object at this position. Since the object is released from rest, its initial kinetic energy is 0. Using the conservation of energy principle, we can equate the total mechanical energy to the sum of the kinetic and potential energy:

Etotal = KE + PE = 0 + U(4.00, 0.30) = -31.99 J.

Since KE = Etotal - PE, we can solve for the speed of the object:

KE = 0.0900 v²/2 = -31.99 J
v² = (2)(-31.99)/(0.0900)
v = ±√(-1122.77) (Note: The negative sign is discarded because speed cannot be negative)
v ≈ ±33.51 m/s

So, the speed of the object when it is at x = 4.00 m and y = 0.30 m is approximately 33.51 m/s.

To find the acceleration of the object at this position, we can use the force corresponding to the potential energy function. The force can be derived from the relation F(x,y) = -dU/dxî - dU/dyĵ.

dU/dx = 2αx = 2(2.00)(4.00) = 16.00 J/m

dU/dy = 3βy² = 3(0.300)(0.30)² = 0.0081 J/m

So, the force corresponding to the potential energy function is F(x,y) = -16.00î + 0.0081ĵ.

Since the only force on the object is conservative, we can use Newton's second law in the form F = ma to find the acceleration:

F = m a
-16.00î + 0.0081ĵ = (0.0900)(xî + yĵ) = 0.0900 xî + 0.0900 yĵ

From this, we can equate the components of the force and acceleration:

-16.00 = 0.0900 x
x = -16.00/0.0900 ≈ -177.78 m
0.0081 = 0.0900 y
y = 0.0081/0.0900 ≈ 0.09 m
So, the acceleration of the object at x = 4.00 m and y = 0.30 m is approximately -177.78î + 0.09ĵ m/s².

User Grr
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