An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+x-xcos(x) #. How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

1 Answer
Jun 29, 2016

Answer:

#= 7g ( 4 pi (1+ 2 pi )) #

Explanation:

the simple idea here is that Work = Force #times# Distance [or more specifically #W = int_C \ vec F * d vec x#]

The work is being done against friction, and here, we have for the friction force....

#F_f = mu R = mu m g = 7g mu#

where #mu(x) = 1+x-xcos x#

so

#F_{f} (x) = 7g (1+x-xcos x) #

The work is being done against friction along a linear path ; and thus...

#W = 7g \ int_{x = 0}^{4 pi} dx qquad \ 1+x- color{red}{xcos x}#

The red term can be done by IBP, ie

#int u v' = uv - int u' v#

We have

#u = x, u' = 1#
#v' = cos x, v = sin x#

so

#int dx qquad x cos x = x sin x - int dx quad sin x#
# = x sin x + cos x#

So

#W = 7g \ int_{x = 0}^{4 pi} dx qquad \ 1+x- xcos x#

#= 7g [x+x^2/2- x sin x - cos x]_{x = 0}^{4 pi}#

#= 7g { [4 pi+(4 pi)^2/2 - 1] - [ -1]}#

#= 7g ( 4 pi (1+(4 pi)/2 ))#

#= 7g ( 4 pi (1+ 2 pi )) \ J#