An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+cotx #. How much work would it take to move the object over #x in [(pi)/4, (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)= 1+cotx   #.  How much work would it take to move the object over #x in [(pi)/4, (7pi)/8], where x is in meters?

Sebastian Acosta · Accepted Answer

The work is #=79.4J#

We need

#intcotxdx=ln|sin(x)|+C#

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(1+cotx)g N#

The work done is

#W=6gint_(1/4pi)^(7/8pi)(1+cotx)dx#

#=6g*[x+ln|sin(x)|]_(1/4pi)^(7/8pi)#

#=6g((7/8pi+ln|sin(7/8pi)|))-(1/4pi+ln|sin(1/4pi)|))#

#=6g(5/8pi-0.614)#

#=6g(1.35)#

#=79.4J#

Benjamin Asiedu · Answer

undefined To find the work done to move the object over the given interval, we need to integrate the force of kinetic friction over the distance $ x $ within the interval $\left[\frac{\pi}{4}, \frac{7\pi}{8}\right]$.

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. However, since the coefficient of kinetic friction is given as $ u_k(x) = 1 + \cot(x) $, we need to find the normal force $ N $ as well.

The normal force $ N $ can be calculated using the gravitational force $ F_g $ acting on the object, which is given by $ F_g = m \cdot g $, where $ m $ is the mass of the object (6 kg) and $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $).

Now, we need to determine the normal force $ N $ at any given position $ x $ along the path. Since the object is being pushed along a linear path, the normal force is equal to the gravitational force acting on the object at any point along the path.

Therefore, the normal force $ N $ is constant and equal to $ m \cdot g $.

Substituting the given expression for the coefficient of kinetic friction $ u_k(x) = 1 + \cot(x) $ and the normal force $ N = m \cdot g $ into the expression for the force of kinetic friction $ F_k = \mu_k \cdot N $, we get:

$$ F_k(x) = (1 + \cot(x)) \cdot (m \cdot g) $$

The work done $ W $ to move the object over the interval $ \left[\frac{\pi}{4}, \frac{7\pi}{8}\right] $ is given by the integral of the force of kinetic friction $ F_k(x) $ with respect to $ x $ over the interval:

$$ W = \int_{\frac{\pi}{4}}^{\frac{7\pi}{8}} F_k(x) \, dx $$

Now, we can substitute the expression for $ F_k(x) $ into the integral and evaluate it over the given interval. However, please note that the integration of $ 1 + \cot(x) $ may require special techniques, such as trigonometric identities or substitution.

After evaluating the integral, we will obtain the work done to move the object over the specified interval.