The force applied against an object moving horizontally on a linear path is described by #F(x)=x^2-x+1 #. By how much does the object's kinetic energy change as the object moves from # x in [ 1, 2 ]#?
Given,
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To find the change in kinetic energy, we need to integrate the force function over the given interval and then apply the work-energy theorem, which states that the work done by the force is equal to the change in kinetic energy.
The work done by the force (F(x)) over the interval ([1, 2]) is given by:
[ W = \int_{1}^{2} F(x) , dx = \int_{1}^{2} (x^2 - x + 1) , dx ]
Evaluating this integral gives:
[ W = \left[ \frac{x^3}{3} - \frac{x^2}{2} + x \right]_{1}^{2} ]
[ W = \left( \frac{2^3}{3} - \frac{2^2}{2} + 2 \right) - \left( \frac{1^3}{3} - \frac{1^2}{2} + 1 \right) ]
[ W = \left( \frac{8}{3} - 2 + 2 \right) - \left( \frac{1}{3} - \frac{1}{2} + 1 \right) ]
[ W = \left( \frac{8}{3} \right) - \left( \frac{1}{3} \right) ]
[ W = \frac{7}{3} ]
The change in kinetic energy is equal to the work done by the force, so:
[ \Delta KE = \frac{7}{3} ]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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