# An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 2+cscx #. How much work would it take to move the object over #x in [pi/12, (5pi)/6]#, where x is in meters?

##### 1 Answer

#### Answer:

#### Explanation:

**Work** done by a *variable force* is defined as the area under the force vs. displacement curve. Therefore, it can be expressed as the integral of force:

#color(darkblue)(W=int_(x_i)^(x_f)F_xdx)# where

#x_i# is the object's initial position and#x_f# is the object's final position

Assuming **dynamic equilibrium**, where we are applying enough force to move the object but not enough to cause a net force and therefore not enough to cause an acceleration, our parallel forces can be summed as:

#sumF_x=F_a-f_k=0#

Therefore we have that

We also have a state of dynamic equilibrium between our perpendicular forces:

#sumF_y=n-F_g=0#

#=>n=mg#

We know that

#vecf_k=mu_kmg#

#=>color(darkblue)(W=int_(x_i)^(x_f)mu_kmgdx)#

**We have the following information:**

#|->"m"=6"kg"# #|->mu_k(x)=2+csc(x)# #|->x in[pi/12,(5pi)/6]# #|->g=9.81"m"//"s"^2#

Returning to our integration, know the

#color(darkblue)(W=mgint_(x_i)^(x_f)mu_kdx)#

Substituting in our known values:

#=>W=(6)(9.81)int_(pi/12)^((5pi)/6)2+csc(x)dx#

Evaluating:

#=>(6)(9.81)[2x+lnabs(csc(x)-cot(x))]_(pi/12)^((5pi)/6)#

#=>(58.86)[((5pi)/3+lnabs(2-(-sqrt3)))-(pi/6+lnabs(csc(pi/12)-cot(pi/12)))]#

#=>(58.86)[(3pi)/2+lnabs(2+sqrt3)-lnabs(csc(pi/12)-cot(pi/12))]#

#=>~~474.23#

Therefore, we have that the work done is