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?
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?
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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?
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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.
