# An object with a mass of #8 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= x^2+3x #. How much work would it take to move the object over #x in [2, 3], where x is in meters?

The work is

We require

The completed work is

The force of friction is

The completed work is

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The work done to move an object against friction can be calculated using the formula:

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

Where:

- ( W ) is the work done (in joules)
- ( F_k(x) ) is the kinetic friction force
- ( x_1 ) and ( x_2 ) are the initial and final positions, respectively

The kinetic friction force ( F_k(x) ) can be determined using the formula ( F_k(x) = \mu_k(x) \cdot N ), where ( \mu_k(x) ) is the kinetic friction coefficient and ( N ) is the normal force.

Given that ( \mu_k(x) = x^2 + 3x ), and the mass of the object is 8 kg, the normal force ( N ) can be calculated as ( N = mg ), where ( g ) is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the values and integrating over the interval ([2, 3]):

[ W = \int_{2}^{3} (x^2 + 3x) \cdot mg , dx ]

[ W = \int_{2}^{3} (x^2 + 3x) \cdot (8 \cdot 9.8) , dx ]

[ W = \int_{2}^{3} (8x^2 + 24x) , dx ]

[ W = \left[ \frac{8}{3}x^3 + 12x^2 \right]_{2}^{3} ]

[ W = \left( \frac{8}{3}(3)^3 + 12(3)^2 \right) - \left( \frac{8}{3}(2)^3 + 12(2)^2 \right) ]

[ W = \left( \frac{8}{3}(27) + 12(9) \right) - \left( \frac{8}{3}(8) + 12(4) \right) ]

[ W = \left( 72 + 108 \right) - \left( \frac{64}{3} + 48 \right) ]

[ W = 180 - \left( \frac{64}{3} + 48 \right) ]

[ W = 180 - \left( \frac{64 + 144}{3} \right) ]

[ W = 180 - \left( \frac{208}{3} \right) ]

[ W = 180 - 69.333 ]

[ W ≈ 110.667 , \text{J} ]

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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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