A small object with mass m = 0.0900 kg moves along the +x- axis. The only force

Question

A small object with mass m = 0.0900 kg moves along the +x- axis. The only force on the object is a conservative force that has the potential-energy function U(x) = -alpha x^2 + betax^3, where alpha = 3.50 J/m^2 and beta = 0.300 J/m^3. The object is released from rest at small x. When the object is at x = 4.00 m, what is its speed? Express your answer with the appropriate units. When the object is at x = 4.00 m, what is the magnitude of its acceleration? Express your answer with the appropriate units.

Explanation / Answer

Here ,

U = - 3.5 x^2 + 0.3 x^3

mass , m = 0.09 Kg

at x = 0

U(0) = 0 J

Using COnservation of energy

change in kinetic energy + change in potential energy = 0

0.5 * m * v^2 = - U(4) + U(0)

0.5 * 0.09 * v^2 = 3.5 * 4^2 - 0.3 * 4^3

solving for v

v = 28.6 m/s

the speed of particle is 28.6 m/s

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B)

for the force applied

F = - dU/dx

F = -d/dx(- 3.5 x^2 + 0.3 x^3)

F = 7 x - 0.9 * x^2

at x = 4 m

F = 7 * 4 - 0.9 * 4^2

F = 13.6 N

Using second law of motion

F = m * a

13.6 = 0.09 * a

a = 151.1 m/s^2

the magnitude of acceleration is 151.1 m/s^2

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