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 [(-5pi)/12, (pi)/6], 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 [(-5pi)/12, (pi)/6], where x is in meters?

Charles Aeschliman · Accepted Answer

The work is #=679.83J#

We require

#intsecxdx=ln(tanx+secx)+C#

The completed work is

#W=F*d#

The force of friction is

#F_r=mu_k*N#

The normal force is #N=mg#

The mass is #m=7kg#

#F_r=mu_k*mg#

#=7*(4+secx)g#

The completed work is

#W=7gint_(-5/12pi)^(1/6pi)(4+secx)dx#

#=7g*[4x+ln(tanx+secx)]_(-5/12pi)^(1/6pi)#

#=7g(4/6pi+ln(tan(pi/6)+sec(pi/6))-(-20/12pi+ln(tan(-5/12pi)+sec(-5/12pi))#

#=7g(9.91)#

#=679.83J#

Parker Abbott · Answer

undefined To calculate the work done to move the object over the given interval, we need to integrate the force of friction with respect to distance. The force of kinetic friction $ F_k $ can be expressed as $ F_k = \mu_k \cdot N $, where $ \mu_k $ is the coefficient of kinetic friction and $ N $ is the normal force.

The normal force $ N $ can be calculated as $ N = mg $, where $ m $ is the mass of the object and $ g $ is the acceleration due to gravity.

Substituting the given values, we have $ N = (7 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) = 68.6 \, \text{N} $.

Now, we need to integrate the force of friction over the given interval $ \left[ -\frac{5\pi}{12}, \frac{\pi}{6} \right] $:

$$ W = \int_{-\frac{5\pi}{12}}^{\frac{\pi}{6}} F_k \, dx $$

$$ W = \int_{-\frac{5\pi}{12}}^{\frac{\pi}{6}} (4 + \sec(x)) \cdot (68.6) \, dx $$

$$ W = 68.6 \int_{-\frac{5\pi}{12}}^{\frac{\pi}{6}} (4 + \sec(x)) \, dx $$

$$ W = 68.6 \left[ 4x + \ln|\sec(x) + \tan(x)| \right]_{-\frac{5\pi}{12}}^{\frac{\pi}{6}} $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} - \frac{5\pi}{6} + \ln\left|\sec\left(\frac{\pi}{6}\right) + \tan\left(\frac{\pi}{6}\right)\right| - \left(-\frac{5\pi}{6} + \ln\left|\sec\left(-\frac{5\pi}{12}\right) + \tan\left(-\frac{5\pi}{12}\right)\right|\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} - \frac{5\pi}{6} + \ln\left|\sqrt{3} + 1\right| - \left(-\frac{5\pi}{6} + \ln\left|\sqrt{3} - 1\right|\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} - \frac{5\pi}{6} + \ln\left(\sqrt{3} + 1\right) + \frac{5\pi}{6} - \ln\left(\sqrt{3} - 1\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} + \ln\left(\sqrt{3} + 1\right) - \ln\left(\sqrt{3} - 1\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} + \ln\left(\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} + \ln\left(\frac{(\sqrt{3} + 1)^2}{3 - 1}\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} + \ln\left(\frac{3 + 2\sqrt{3} + 1}{2}\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} + \ln\left(\frac{4 + 2\sqrt{3}}{2}\right) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} + \ln(2 + \sqrt{3}) \right] $$

$$ W \approx 68.6 \left[ \frac{\pi}{3} + 0.916 \right] $$

$$ W \approx 68.6 \times 1.926 $$

$$ \boxed{W \approx 132.159 \, \text{J}} $$

So, it would take approximately 132.159 Joules of work to move the object over the given interval.