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The force applied against an object moving horizontally on a linear path is described by #F(x)=x^2+ 5 #. By how much does the object's kinetic energy change as the object moves from # x in [ 1, 2 ]#?

Answer 1

Julia Aguilar

The change in kinetic energy is #=7.33J#

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

The change in kinetic energy is equal to the work.

#DeltaW=F(x)Deltax#

Therefore,

#W=int_1^2(x^2+5)dx#

#=[x^3/3+5x]_1^2#

#=(8/3+10)-(1/3+5)#

#=5+7/3#

#=7.33J#

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Answer 2

Sebastian Acosta

The change in kinetic energy as the object moves from ( x = 1 ) to ( x = 2 ) is ( \Delta KE = \int_{1}^{2} F(x) , dx ). Given ( F(x) = x^2 + 5 ), integrating ( F(x) ) from ( x = 1 ) to ( x = 2 ) gives ( \Delta KE = \left[ \frac{x^3}{3} + 5x \right]_{1}^{2} ). Evaluating this yields ( \Delta KE = \left( \frac{2^3}{3} + 5 \cdot 2 \right) - \left( \frac{1^3}{3} + 5 \cdot 1 \right) = \frac{8}{3} + 10 - \frac{1}{3} - 5 = \frac{7}{3} + 5 = \frac{22}{3} ) units.

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Answer from HIX Tutor

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.