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

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Answer 1 · 2016-07-14T13:21:41

#-49.5J#

The change in KE is work done against varible frictional force.

So

#-int_2^5F(x)dx#

#=-int_2^5(x^2+x)dx#

#=-[x^3/3+x^2/2]_2^5#

#=-(5^3/3+5^2/2-2^3/3-2^2/2)#

#=-(117/3+21/2)J#

#=-(39+10.5)=-49.5J#

#color(red)("Negative sign means work done against the force")#

#color(red)(",representing decrease in KE")#