Question about maximum Kinetic Energy? KE & Work problem?

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The discussion centers on calculating the maximum kinetic energy (KE) of a 1.5 kg block subjected to a variable force F(x) = (2.5 - x²)i N on a frictionless surface. The maximum kinetic energy occurs at x = 1.6 m, where the force F becomes zero, indicating a transition from positive to negative force. This is derived from the principle that kinetic energy is maximized when the net force acting on the object is zero, as it signifies the point where the object stops accelerating and begins to decelerate.

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nchin
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Question about maximum Kinetic Energy?? KE & Work problem??

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along an x-axis
is applied to the block. The force is given by F(x) = (2.5 - x^(2))i N, where x is in meters and the initial position of the
block is x = 0.

a) What is the kinetic energy of the block as it passes through x = 2.0 m ?

b) What is the maximum kinetic energy of the block between x = 0 and x = 2.0 m ?

I understand how to solve a but its part b that confuses me.

Solution Guide states:
"KE is maximized when F = 0.
F = 0 when 2.5 - x^(2) = 0
x = √ 2.5 = 1.6 m
Thus KE is maximized at x = 1.6 m "

But why is KE maximized when F = 0?? Someone please explain to me.

Thanks!
 
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nchin said:
b) What is the maximum kinetic energy of the block between x = 0 and x = 2.0 m ?

But why is KE maximized when F = 0??

So I assume you did (a) by integrating F(x) over [0,2]. This should give you the intuition that [itex]Work = \int F(x) = \Delta KE = KE[/itex]

Now to find a maximum, you simply need to take the derivative and set = to 0. If take the derivative of KE, we're back to F(x)!

You'll notice that when x > 1.6, F(x) is negative. This means that once we're past x=1.6, the force becomes negative, or starts pulling instead of pushing. Of course, this subtracts from the KE of the block. In short, the maximum KE occurs at the instant the force goes from positive to negative.
 

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