An object with a mass of #1 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= e^x-1 #. How much work would it take to move the object over #x in [1, 3], where x is in meters?

Question

An object with a mass of #1 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= e^x-1   #.  How much work would it take to move the object over #x in [1, 3], where x is in meters?

Isabella Alvarez · Accepted Answer

The work is #=150.63J#

#"Reminder : "#

#inte^xdx=e^x+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=(e^x-1)#

The normal force is #N=mg#

The mass of the object is #m=1kg#

#F_r=mu_k*mg#

#=1*(e^x-1)g#

The work done is

#W=1gint_1^3(e^x-1)dx#

#=1g*[e^x-x]_1^3#

#=1g(e^3-3)-(e-1)#

#=1g(15.37)#

#=150.63J#

Charles Aeschliman · Answer

undefined To find the work done to move the object over the given interval [1, 3], you need to integrate the product of the friction force and the displacement over that interval. The work done $ W $ is given by the integral of the friction force $ F_{friction} $ over the displacement $ x $:

$$ W = \int_{1}^{3} F_{friction}(x) \, dx $$

Given that the kinetic friction coefficient $ \mu_k(x) = e^x - 1 $, the friction force $ F_{friction}(x) $ is:

$$ F_{friction}(x) = \mu_k(x) \cdot m \cdot g $$

where $ m $ is the mass of the object and $ g $ is the acceleration due to gravity. Substituting the given values and integrating over the interval [1, 3], you can calculate the work done.