The force applied against a moving object travelling on a linear path is given by #F(x)= cosx + 2 #. How much work would it take to move the object over #x in [ 0, (13 pi) / 6] #?

1 Answer
Jul 22, 2017

Answer:

The work is #=14.1J#

Explanation:

#intcosxdx=sinx+C#

#int(x^n)dx=x^(n+1)/(n+1)+C(n!=-1)#

The work is

#DeltaW=F(x)*Deltax#

#F(x)=2+cosx#

#W=int_0^(13/6pi)(2+cosx)dx#

#=[2x+sinx]_0^(13/6pi)#

#=(2*13/6pi+sin(13/6pi))-(0+sin0)#

#=13/3pi+sin(13/6pi)#

#=31/3pi+1/2#

#=14.1J#