An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+5cotx#. How much work would it take to move the object over #x in [(5pi)/12, (5pi)/8]#, where #x# is in meters?

Answer 1

The work is #=25.28J#

#"Reminder : "#
#intcotxdx=ln(|sinx|)+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=(1+5cot(x))#
The normal force is #N=mg#
The mass of the object is #m=6kg#
#F_r=mu_k*mg#
#=6*(1+5cot(x))g#

The work done is

#W=6gint_(5/12pi)^(5/8pi)(1+5cot(x))dx#
#=6g*[x+5ln(|sinx|)]_(5/12pi)^(5/8pi)#
#=6g(5/8pi+5ln(|sin(5/8pi)|))-(5/12pi+5ln(|5/12pi|)))#
#=6g(*0.43)#
#=25.28J#
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Answer 2

To find the work done to move the object over the given interval, you need to integrate the product of the force of kinetic friction and the displacement over that interval. The force of kinetic friction ( F_k ) is given by ( F_k = \mu_k \cdot N ), where ( \mu_k ) is the coefficient of kinetic friction and ( N ) is the normal force.

Given ( \mu_k(x) = 1 + 5 \cot(x) ), the normal force ( N ) can be calculated from the weight of the object. Since the object is on a linear path, the normal force is equal to the gravitational force acting on it, which is ( mg ), where ( m ) is the mass and ( g ) is the acceleration due to gravity (( g \approx 9.8 , \text{m/s}^2 )).

So, ( N = mg = 6 \times 9.8 = 58.8 , \text{N} ).

The work done ( W ) is given by the integral of the product of ( F_k ) and ( dx ) over the interval ( \left[\frac{5\pi}{12}, \frac{5\pi}{8}\right] ):

[ W = \int_{\frac{5\pi}{12}}^{\frac{5\pi}{8}} F_k , dx ]

[ W = \int_{\frac{5\pi}{12}}^{\frac{5\pi}{8}} (1 + 5 \cot(x)) \cdot N , dx ]

[ W = \int_{\frac{5\pi}{12}}^{\frac{5\pi}{8}} (1 + 5 \cot(x)) \cdot 58.8 , dx ]

After integrating this expression over the given interval, you will get the work done in 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.

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.

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.

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.

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