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 [(pi)/12, (3pi)/8]#, where #x# is in meters?

Answer 1

The work is #=428.1J#

#"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+5cotx)#
The normal force is #N=mg#
The mass of the object is #m=6kg#
#F_r=mu_k*mg#
#=6*(1+5cotx)g#

The work done is

#W=6gint_(1/12pi)^(3/8pi)(1+5cot)dx#
#=6g*[x+5ln(sin(x) ] _(1/12pi)^(3/8pi)#
#=6g((3/8pi+5ln(sin(3/8pi))-(1/12pi+5ln(sin(1/12pi))#
#=6xx9.8xx7.28#
#=428.1J#
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Answer 2

To find the work done in moving the object over the given interval, we need to integrate the product of the force of kinetic friction and the distance 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.

The normal force ( N ) is equal to the weight of the object, which is ( m \cdot g ), where ( m ) is the mass of the object and ( g ) is the acceleration due to gravity.

Given ( m = 6 ) kg, ( g \approx 9.8 ) m/s(^2), and the interval for ( x ) is ( \left[\frac{\pi}{12}, \frac{3\pi}{8}\right] ), we can calculate the work done using the following steps:

  1. Calculate the normal force ( N = m \cdot g ).
  2. Calculate the force of kinetic friction ( F_k = \mu_k \cdot N ) using ( \mu_k(x) = 1 + 5 \cot(x) ).
  3. Integrate the product of ( F_k ) and ( dx ) over the given interval.

[ W = \int_{\frac{\pi}{12}}^{\frac{3\pi}{8}} F_k(x) , dx ]

After calculating ( W ) using the definite integral, you will find the work done to move the object over the specified interval.

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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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