The force applied against a moving object travelling on a linear path is given by #F(x)=x+2e^x #. How much work would it take to move the object over #x in [0,2 ] #?

1 Answer
Jan 17, 2016

Answer:

Work Done #=2e^2or approx 14.78J#

Explanation:

Work done by a Force #vecF# when it moves the object through a distance #vecdS# is

#dW=vecF.vecdS=|vecF|.|vecdS|.cos theta#

Where #theta# is the angle between the two vectors

As the Force is being applied on the object moving in a linear path the angle between the two is #0#, an therefore #costheta=1#

#dW=|vecF|.|vecdS|.#

When the force #F(x)# moves through a small distance #dx#
#dW=F(x)dx#
Total work done to move the object from #x in [0,2]# is integral of #dW#over limits #0# to #2#.

#W=int_0^2(x+2e^x)dx#
#=x^2//2+2e^x+C|_0^2#, where C is constant of integration.
#=(2^2//2+2e^2+C)-(0^2//2+2e^0+C)#
#=(cancel (color (red)2)+2e^2+cancel C)-((cancel (color (red)2)+cancelC)#
#=2e^2#