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, (3pi)/4], 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, (3pi)/4], where x is in meters?

Amelia Adams · Accepted Answer

The work is #=377.5J#

#"Reminder : "#

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

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The coefficient of kinetic friction is #mu_k=(2+csc(x))#

The normal force is #N=mg#

The mass of the object is #m=6kg#

#F_r=mu_k*mg#

#=6*(2+csc(x))g#

The work done is

#W=6gint_(1/8pi)^(3/4pi)(2+csc(x))dx#

#=6g*[2x+ln|(tan(x/2))|]_(1/8pi)^(3/4pi)#

#=6g(3/2pi+ln(tan(3/8pi)))-(1/4pi+3ln(tan(1/16pi))#

#=6g(6.42)#

#=377.5J#

Axel Acevedo · Answer

undefined To calculate the work done to move the object over the given interval, you need to integrate the force of kinetic friction with respect to distance $ x $ over the interval $ [\frac{\pi}{8}, \frac{3\pi}{4}] $.

The force of kinetic friction is given by $ f_k(x) = \mu_k(x) \cdot N $, where $ \mu_k(x) = 2 + \csc(x) $ and $ N $ is the normal force.

Given that the object is pushed along a linear path, the normal force is constant. Let's denote it as $ N_0 $.

Therefore, the work done is given by the integral:

$$ W = \int_{\frac{\pi}{8}}^{\frac{3\pi}{4}} f_k(x) \, dx $$

$$ W = \int_{\frac{\pi}{8}}^{\frac{3\pi}{4}} (2 + \csc(x)) \cdot N_0 \, dx $$

You'll need to evaluate this integral to find the exact amount of work done.