An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 4+secx #. How much work would it take to move the object over #x in [0, (pi)/12], where x is in meters?

1 Answer
Apr 13, 2017

Answer:

The work is #=89.7J#

Explanation:

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We need

#intsecxdx=ln(tanx+secx)#

The frictional force is

#F=F_r=mu_k*N#

The normal force is #N=mg#

So,

#F_r=mu_k*mg#

But,

#mu_k(x)=4+secx#

The work done is

#W=F*d=int_0^(pi/12)mu_kmgdx#

#=7gint_0^(pi/12)(4+secx)dx#

#=7g[4+ln(tanx+1/cosx)]_0^(pi/12)#

#=7g((pi/3+0.26)-(0))#

#=7g*1.31#

#=89.7J#