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

1 Answer
Jan 18, 2016

Answer:

Work #=269.5J#

Explanation:

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]#