# An object with a mass of #4 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 5+tanx #. How much work would it take to move the object over #x in [(-5pi)/12, (5pi)/12], where x is in meters?

The work is

The completed work is

[5x+ln|cos(x)|]_(-5/12pi)^(5/12pi)# = 4g*

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[W = \int_{-5\pi/12}^{5\pi/12} \left(\mu_k(x) \cdot m \cdot g + \frac{m \cdot v^2(x)}{2}\right) , dx]

[W = \int_{-5\pi/12}^{5\pi/12} \left((5 + \tan(x)) \cdot 4 \cdot 9.8 + \frac{4 \cdot v^2(x)}{2}\right) , dx]

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To find the work done to move the object over the given interval, we integrate the force of kinetic friction with respect to displacement (x) over the specified range. The work done by friction is given by the formula:

[ W = \int_{x_1}^{x_2} f_k(x) , dx ]

Given that the coefficient of kinetic friction ( \mu_k(x) = 5 + \tan(x) ), we can find the force of kinetic friction ( f_k(x) ) using the formula:

[ f_k(x) = \mu_k(x) \cdot N ]

where N is the normal force. Since the object is pushed along a linear path, the normal force N is equal to the gravitational force acting on the object, which is ( mg ), where ( m = 4 , \text{kg} ) is the mass of the object and ( g = 9.8 , \text{m/s}^2 ) is the acceleration due to gravity.

So, ( N = mg = 4 \times 9.8 = 39.2 , \text{N} ).

Therefore, ( f_k(x) = (5 + \tan(x)) \times 39.2 ).

Now, we integrate ( f_k(x) ) with respect to ( x ) over the interval ( \left[ \frac{-5\pi}{12}, \frac{5\pi}{12} \right] ):

[ W = \int_{-\frac{5\pi}{12}}^{\frac{5\pi}{12}} (5 + \tan(x)) \cdot 39.2 , dx ]

After finding this integral, you'll have the work done to move the object over the specified interval.

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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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