SOLUTION: A potential function is given by U(x) = 15 x^2. What will be the acceleration (in meters/second ^2) of a 2.00 kg mass, when it is at the position x = 0.300 m?

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Question 1056860: A potential function is given by U(x) = 15 x^2. What will be the acceleration (in meters/second ^2) of a 2.00 kg mass, when it is at the position x = 0.300 m?

Answer by ikleyn(52790)   (Show Source): You can put this solution on YOUR website!
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A potential function is given by U(x) = 15 x^2. What will be the acceleration (in meters/second ^2) of a 2.00 kg mass,
when it is at the position x = 0.300 m?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In a potential field U(x) the force acting to the mass "m" is

F = .

In our case this force is F = -m*(30x) = -30mx.

Hence, the acceleration is

a =  = -30x.

Substitute your data:

a = -30*0.300 = -9 .

The sign "-" means that the force and the acceleration are directed to the origin of the coordinate system.

I hope that the expression for U(x) is given in appropriate units.


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