An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+x^2-xcos(x) #. How much work would it take to move the object over #x in [pi, 4pi], 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)= 1+x^2-xcos(x)   #.  How much work would it take to move the object over #x in [pi, 4pi], where x is in meters?

Nora Arzate · Accepted Answer

The work done is #=45.18kJ#

#"Reminder : "#

Integration between components

#intuv'=uv-intu'v#

#u=x#, #=>#, #u'=1#

#v'=cosx#, #=>#, #v=sinx#

#intxcosxdx=xsinx+cosx+C#

The completed work is

#W=F*d#

The force of friction is

#F_r=mu_k*N#

The coefficient of kinetic friction is #mu_k=(1+x^2-xcos(x))#

The normal force is #N=mg#

The mass of the object is #m=7kg#

#F_r=mu_k*mg#

#=7*(1+x^2-xcos(x))g#

The completed work is

#W=7gint_(pi)^(4pi)(1+x^2-xcos(x))dx#

#=7g*[x+1/3x^3-xsin(x)-cosx]_(pi)^(4pi)#

#=7g(4pi+64/3pi^3-4pisin(4pi)-cos(4pi))-(pi+1/3pi^3+1))#

#=7g(-2+3pi+21pi^3)#

#=45177.2J#

Sebastian Acosta · Answer

undefined To calculate the work done to move the object over the given interval, we need to integrate the product of the force of friction and the distance traveled with respect to x:

$$ W = \int_{\pi}^{4\pi} F_{\text{friction}}(x) \, dx $$

Given that $ F_{\text{friction}}(x) = \mu_k(x) \cdot N $, where $ N $ is the normal force, and $ \mu_k(x) = 1 + x^2 - x\cos(x) $, we can find the normal force as $ N = mg $, where $ m $ is the mass of the object (7 kg) and $ g $ is the acceleration due to gravity (9.8 m/s²).

Now, we can integrate:

$$ W = \int_{\pi}^{4\pi} (1 + x^2 - x\cos(x)) \cdot mg \, dx $$

$$ W = \int_{\pi}^{4\pi} (1 + x^2 - x\cos(x)) \cdot (7 \times 9.8) \, dx $$

$$ W = 68 \int_{\pi}^{4\pi} (1 + x^2 - x\cos(x)) \, dx $$

After performing the integration over the given interval, we will obtain the work done to move the object over that distance.