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

Answer 1

#W = 1236# #"J"#

We're asked to find the necessary work that needs to be done on a #3#-#"kg"# object to move on the position interval #x in [1color(white)(l)"m", 3color(white)(l)"m"]# with a varying coefficient of kinetic friction #mu_k# represented by the equation
#ul(mu_k(x) = 3x^2-2x+12#
The work #W# done by the necessary force is given by
#ulbar(|stackrel(" ")(" "W = int_(x_1)^(x_2)F_xdx" ")|)# #color(white)(a)# (one dimension)

where

#F_x# is the magnitude of the necessary force
#x_1# is the original position (#1# #"m"#)
#x_2# is the final position (#3# #"m"#)

The necessary force would need to be equal to (or greater than, but we're looking for the minimum value) the retarding friction force, so

#F_x = f_k = mu_kn#
Since the surface is horizontal, #n = mg#, so
#ul(F_x = mu_kmg#
The quantity #mg# equals
#mg = (3color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 29.43color(white)(l)"N"#

And we also plug in the above coefficient of kinetic friction equation:

#ul(F_x = (29.43color(white)(l)"N")(3x^2-2x+12)#
Work is the integral of force, and we're measuring it from #x=1# #"m"# to #x=3# #"m"#, so we can write
#color(red)(W) = int_(1color(white)(l)"m")^(3color(white)(l)"m")(29.43color(white)(l)"N")(3x^2-2x+12) dx = color(red)(ulbar(|stackrel(" ")(" "1236color(white)(l)"J"" ")|)#
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Answer 2

We observe that Coefficient of kinetic friction is given as

#u_k(x)=3x^2−2x+12#

and that the object is being pushed along a linear path.

We know that force of friction opposes the motion. As such we can infer that movement is only in the #x#-direction. The angle between force of friction and displacement is #180^@#
Work Done against force #vecF# for moving a distance #vecs# is given as
#W=vecFcdotvecs#
Let the object move a distanace #vecdx#. Work done to move the object against the force of friction
#dW=vec(F_f)cdot vecdx#
Force of friction #vec(F_f)=-u_kNhatx# where #N# is normal reaction and given #=mg# #dW=-(3x^2−2x+12)mghatxcdot vecdx#
Integrating both sides for the given interval we get #W=-int_1^3(3x^2−2x+12)mgdx#
#W=-3xx9.81|3x^3/3−2x^2/2+12x|_1^3# #W=-29.43[(3^3−3^2+12xx3)-(1^3−1^2+12xx1)]# #W=-29.43[(27−9+36)-(1−1+12)]# #W=-29.43(54-12)# #W=-1236J#
#-ve# sign shows that this amount of energy was lost in doing work against friction.
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Answer 3

The work required to move the object over ( x ) in [1, 3] is approximately 62.33 joules.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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