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

Question

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

Isabella Alvarez · Accepted Answer

#W = 335# #"J"#

We're asked to find the necessary work that needs to be done on a #5#-#"kg"# object to move on the position interval #x in [2color(white)(l)"m", 3color(white)(l)"m"]# with a varying coefficient of kinetic friction #mu_k# represented by the equation

#ul(mu_k(x) = x^2-x+3#

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 (#2# #"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 = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 49.05color(white)(l)"N"#

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

#ul(F_x = (49.05color(white)(l)"N")(x^2-x+3)#

Work is the integral of force, and we're measuring it from #x=3# #"m"# to #x=4# #"m"#, so we can write

#color(red)(W) = int_(2color(white)(l)"m")^(3color(white)(l)"m")(49.05color(white)(l)"N")(x^2-x+3) dx = color(red)(ulbar(|stackrel(" ")(" "335color(white)(l)"J"" ")|)#

Adam Arnold · Answer

undefined To calculate the work done against kinetic friction, use the integral of the friction force over the given interval:

$$ W = \int_{2}^{3} \mu_k(x) \cdot m \cdot g \,dx $$

where $ \mu_k(x) $ is the kinetic friction coefficient, $ m $ is the mass of the object (5 kg), $ g $ is the acceleration due to gravity (approximately 9.8 m/s²).

$$ W = \int_{2}^{3} (x^2 - x + 3) \cdot 5 \cdot 9.8 \,dx $$

Evaluate the definite integral over the interval [2, 3].