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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] #?

Answer 1

Emily Abbate

The work is #=14.1J#

#intcosxdx=sinx+C#

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

The task 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#

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Answer 2

Nicholas Agrusa

To calculate the work done, integrate the force function F(x) with respect to x over the given interval [0, (13π)/6]:

∫[0, (13π)/6] (cos(x) + 2) dx

= [sin(x) + 2x] evaluated from 0 to (13π)/6

= (sin((13π)/6) + 2(13π)/6) - (sin(0) + 2(0))

= (-(1/2) + (13π)/3) - (0 + 0)

= -(1/2) + (13π)/3

So, the work done to move the object over the interval [0, (13π)/6] is -(1/2) + (13π)/3.

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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.