An object with a mass of #8 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 2x^2-3x #. How much work would it take to move the object over #x in [2, 3], where x is in meters?

1 Answer
Mar 20, 2016

Answer:

#797.1J# rounded to one place of decimal

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#
In the problem #mu_k# is stated to be #u_k(x)#
Inserting given quantities and taking the value of #g=9.8 m//s^2#
#F_k=(2x^2-3x)times 8times 9.8 N#
#F_k=78.4(2x^2-3x) N#

When this force moves through a small distance #dx#, the work done is given as
#F_k cdotdx=[78.4(2x^2-3x)]cdot 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 78.4(2x^2-3x)cdotdx#
#implies #Total work done#=78.4 int_2^3 (2x^2-3x)cdotdx#

#=78.4 (2x^3 /3-3x^2/2+C)|_2^3#, where C is constant of integration.
#=78.4 [ (2 3^3 /3-3 3^2/2+cancel C)-(2 2^3 /3-3 2^2/2+cancel C)]#
#=78.4 [ 18-27/2-16/3+6]#
#=78.4 [ (108-51-32+36)/6]#
#=78.4 [ 61/6]#
#797.1J# rounded to one place of decimal