A 2.3 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F with arrow(x) = (2.6 − x^2) 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 = 2.1 m?

(b) What is the maximum kinetic energy of the block between x = 0 and x = 2.1 m?

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(a) K naught should tacitly be equal to 0, so when finding W, we should in essence find the final K.
F(x=2.1m)=(2.6-(2.1)^2) = -1.81N
If I plug -1.81N for the scalar product or use integration, I get -3.80J. This is incorrect however.

(b) Finding the max K in (0, 2.1] should be related to finding the zeros of its derivative, since that indicates either a max or min in the interval. The only solution for F'(x) is 0, however.

I am confused..

Any redirection or indication of screw-ups is greatly appreciated.

integration is the correct approach

work (energy) = F(x) dx

the force goes to zero at √2.6, and then goes negative
... this is also the point of max energy

I apologize for the confusion. Let's go through the problem step by step to find the correct solutions.

(a) To find the kinetic energy of the block as it passes through x = 2.1 m, we need to calculate the work done on the block from x = 0 to x = 2.1 m, and then equate it to the change in kinetic energy.

The work done is given by:
W = ∫ F(x) dx

Using the equation for the force, F(x) = 2.6 - x^2, the work done becomes:
W = ∫ (2.6 - x^2) dx

Evaluating the integral, we get:
W = [2.6x - (1/3)x^3] evaluated from x = 0 to x = 2.1

W = 2.6(2.1) - (1/3)(2.1)^3 - (2.6(0) - (1/3)(0)^3)
W = 5.46 - (1/3)(9.261) - 0
W ≈ 5.46 - 3.087
W ≈ 2.373 J

Since the block is initially at rest, its initial kinetic energy is 0. Therefore, the kinetic energy of the block as it passes through x = 2.1 m is approximately 2.373 J.

(b) To find the maximum kinetic energy of the block between x = 0 and x = 2.1 m, we need to find the maximum value of the function F(x) = 2.6 - x^2 within this interval.

To find the maximum, we can take the derivative of F(x) with respect to x, and set it equal to 0:
F'(x) = -2x = 0

Solving for x, we get:
x = 0

Since x = 0 is the only solution, it represents the maximum value of F(x) in the interval (0, 2.1].

Therefore, the maximum kinetic energy of the block between x = 0 and x = 2.1 m occurs at x = 0, and it is 2.6 J (as there is zero work done, so all the applied force is converted to kinetic energy).

I hope this clarifies the solution. Let me know if you have any further questions!

Let's go through the problem step by step to find the correct answers.

(a) To find the kinetic energy of the block at x = 2.1 m, we need to calculate the work done on the block by the force F(x) from x = 0 to x = 2.1 m. The work done on an object is given by the equation:

W = ∫F(x)dx

The equation provided for F(x) is F(x) = 2.6 N - x^2 N. We can plug this into the work equation:

W = ∫(2.6 - x^2)dx

Integrating this equation will give us the work done on the block. However, since the equation for F(x) is given only up to x = 2.1 m, we need to evaluate the integral from x = 0 to x = 2.1 m:

W = ∫(2.6 - x^2)dx evaluated from 0 to 2.1

Evaluating the integral and solving for W will give us the work done on the block. The kinetic energy of the block is equal to the work done on it. So, the kinetic energy of the block at x = 2.1 m is equal to the obtained value of W.

(b) To find the maximum kinetic energy of the block between x = 0 and x = 2.1 m, we need to find the maximum value of the kinetic energy function K(x) in that interval. The kinetic energy of an object is given by:

K(x) = ∫F(x')dx' evaluated from 0 to x

To find the maximum value, we need to find the derivative of K(x) and solve for when it equals zero. This will give us the locations of potential maxima or minima in the interval (0, 2.1]. The maximum kinetic energy will occur at the maximum point.

So, to find the maximum kinetic energy, we can differentiate the kinetic energy function K(x) and solve for its derivative equals zero. Then, substitute the values obtained into the kinetic energy function to get the maximum kinetic energy.

I hope this clarification helps you understand how to proceed with solving the problem. If you have any further questions or need more assistance, please let me know.