An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 6x-1 #. How much work would it take to move the object over #x in [2, 3], where x is in meters?

1 Answer
Jan 19, 2016

Answer:

See details in Explanation section

Explanation:

The question is similar to another question solved as bellow. The difference is in the value of coefficient of friction. Choose appropriate value and solve.
In case of difficulty, use comment section below.
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Force of kinetic friction which needs to be overcome to move the object

#F_k=#Coefficient of kinetic friction #mu_ktimes #normal force #eta#
where #eta=mg#
Inserting given quantities and taking the value of #g=9.8 m//s^2#
#F_k=(x+3)times 5times 9.8 N#
#F_k=49(x+3) N#

When this force moves through a small distance #dx#, the work done is given as
#F_k dx=49(x+3) dx#
When the force moves through a distance from #x in [2, 3]#, total work done is integral of RHS over the given interval.

Total work done#=int_2^3 49(x+3)dx#
#implies #Total work done#=49 int_2^3 (x+3)dx#

#=49 (x^2 /2+3x+C)|_2^3#, where C is constant of integration.
#=49 [ (3^2 /2+3times3+C)-(2^2 /2+3times2+C)]#
#=49 [ 9 /2+9-2-6]#
#=49 [ 4.5+1]#