# The force applied against an object moving horizontally on a linear path is described by #F(x)= 4x^2+x #. By how much does the object's kinetic energy change as the object moves from # x in [ 1, 2 ]#?

Writing the acceleration in terms of the force is how we start.

then combine the two sides

As is well known, the kinetic energy is provided by

Delta K is equal to (4/3 2^3 + 1/2 2^2+v_1)-(4/3 1^3 + 1/2 1^2+v_1), or 65/6.

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Using the idea of work—which is the force integrated over the applied distance—I've determined that this is actually much simpler than the previous response.

The work we do ultimately causes the object's kinetic energy to increase, so

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The change in kinetic energy of the object as it moves from x = 1 to x = 2 is equal to the work done by the force F(x) over that interval. Therefore, to find the change in kinetic energy, we need to calculate the work done by integrating the force function F(x) with respect to x over the interval [1, 2]. This can be expressed as:

∆KE = ∫[1 to 2] F(x) dx

Where F(x) = 4x^2 + x.

Then, ∆KE = ∫[1 to 2] (4x^2 + x) dx.

After integrating and evaluating the expression, we'll get the change in kinetic energy.

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