An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 4+secx #. How much work would it take to move the object over #x in [0, (pi)/12], where x is in meters?

Question

An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 4+secx   #.  How much work would it take to move the object over #x in [0, (pi)/12], where x is in meters?

Leo Abeyta · Accepted Answer

The work is #=89.7J#

We need

#intsecxdx=ln(tanx+secx)#

The frictional force is

#F=F_r=mu_k*N#

The normal force is #N=mg#

So,

#F_r=mu_k*mg#

But,

#mu_k(x)=4+secx#

The work done is

#W=F*d=int_0^(pi/12)mu_kmgdx#

#=7gint_0^(pi/12)(4+secx)dx#

#=7g[4+ln(tanx+1/cosx)]_0^(pi/12)#

#=7g((pi/3+0.26)-(0))#

#=7g*1.31#

#=89.7J#

Adrian Ayala · Answer

undefined To find the work done to move the object over the given interval, integrate the product of the force of kinetic friction (μ_k) and the distance moved (x) with respect to x over the interval [0, π/12]:

$$ W = \int_{0}^{\frac{\pi}{12}} \mu_k(x) \cdot x \, dx $$

First, find the expression for μ_k(x) by substituting the given function:

$$ \mu_k(x) = 4 + \sec(x) $$

Now, integrate the expression over the given interval:

$$ W = \int_{0}^{\frac{\pi}{12}} (4 + \sec(x)) \cdot x \, dx $$

Once you integrate this expression, you'll get the work done to move the object over the given interval.