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An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1-cos(x/3) #. How much work would it take to move the object over #x in [0, 8pi], where x is in meters?

Answer 1

Leo Abeyta

The work is #=1104.2J#

The completed work is

#W=F*d#

The force of friction is

#F_r=mu_k*N#

#N=mg#

#F_r=mu_k*mg#

#=5(1-cos(x/3))g#

The completed work is

#W=5gint_(0)^(8pi)(1-cos(x/3))dx#

#=5g*[x-3sin(x/3)]_(0)^(8pi)#

#=5g((8pi-3sin(8/3pi))-(0-0))#

#=5g(8pi-3/2sqrt3)#

#=1104.2J#

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Answer 2

Carson Alexander

[ W = \int_{0}^{8\pi} \left( \mu_k \cdot m \cdot g \cdot \cos\left(\frac{x}{3}\right) + m \cdot g \cdot \sin\left(\frac{x}{3}\right) \right) , dx ]

[ W \approx 2550 , \text{Joules} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.