An object with a mass of #1 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 3x^2+12 #. How much work would it take to move the object over #x in [1, 3], where x is in meters?

1 Answer

Answer:

Work #W=490" "#Joules

Explanation:

Given:
mass #m=1 " ""kg"#
Variable kinetic friction coefficient #mu_k(x)=3x^2+12#

Let #x# the distance moved by the object

Let #F_k# the force required to move the object
Let #F_n# the normal force exerted by the surface on the object
#F_n=m*g#

#F_k=mu_k*F_n#
#F_k=(3x^2+12)*m*g#

The amount of work

#dW=F_k*dx#

#W=int_1^3 F_k*dx#

#W=int_1^3 (3x^2+12)*m*g*dx#

#W=mg*int_1^3 (3x^2+12)*dx#

#W=mg[(x^3+12x)]_1^3#

#W=50*mg#

#W=50*(1)*(9.8)#

#W=490" "#Joules

God bless...I hope the explanation is useful