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/8, (7pi)/8], where x is in meters?

Question

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/8, (7pi)/8], where x is in meters?

William Audy · Accepted Answer

The work is #=467J#

We need

#intcscxdx=ln|tan(x/2)|#

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The normal force is #N=mg#

The mass is #m=6kg#

#F_r=mu_k*mg#

#=6(2+cscx)g#

The work done is

#W=6gint_(1/8pi)^(7/8pi)(2+cscx)dx#

#=6g*[2x+ln|tan(x/2)|]_(1/8pi)^(7/8pi)#

#=6g((2*7/8pi+ln|tan(7/16pi)|)-(2*1/8pi+ln|tan(1/16pi)|))#

#=6g(7/4pi-1/4pi+3.22)#

#=6g(3/2pi+3.22)#

#=467J#

Axel Acevedo · Answer

undefined To calculate the work done to move the object over the given interval [π/8, (7π)/8], we need to integrate the force of kinetic friction over this interval. The formula for work done against friction is:

$$ W = \int_{x_1}^{x_2} F_{\text{friction}} \, dx $$

Given that the kinetic friction coefficient $ u_k(x) = 2 + \csc(x) $, we can find the force of kinetic friction $ F_{\text{friction}} $ using the formula:

$$ F_{\text{friction}} = u_k(x) \times \text{normal force} $$

The normal force equals the weight of the object, which is $ mg $, where $ m $ is the mass of the object and $ g $ is the acceleration due to gravity ($ g \approx 9.8 \, \text{m/s}^2 $).

So, we have:

$$ F_{\text{friction}} = (2 + \csc(x)) \times mg $$

Substituting the given mass ($ m = 6 $ kg), and integrating over the interval [π/8, (7π)/8], we get:

$$ W = \int_{\frac{\pi}{8}}^{\frac{7\pi}{8}} (2 + \csc(x)) \times 6 \times 9.8 \, dx $$

$$ W = 6 \times 9.8 \int_{\frac{\pi}{8}}^{\frac{7\pi}{8}} (2 + \csc(x)) \, dx $$

$$ W = 6 \times 9.8 \left[ 2x - \ln|\csc(x) + \cot(x)| \right]_{\frac{\pi}{8}}^{\frac{7\pi}{8}} $$

$$ W \approx 6 \times 9.8 \left[ \left( 2\frac{7\pi}{8} - \ln|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)| \right) - \left( 2\frac{\pi}{8} - \ln|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)| \right) \right] $$

$$ W \approx 6 \times 9.8 \left[ \left( \frac{7\pi}{4} - \ln|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)| \right) - \left( \frac{\pi}{4} - \ln|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)| \right) \right] $$

$$ W \approx 6 \times 9.8 \left[ \left( \frac{7\pi}{4} - \ln\left|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)\right| \right) - \left( \frac{\pi}{4} - \ln\left|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)\right| \right) \right] $$

$$ W \approx 6 \times 9.8 \left( \frac{7\pi}{4} - \ln\left|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)\right| - \frac{\pi}{4} + \ln\left|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)\right| \right) $$