An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 6 x - 1$ $u_k(x)= 6x-1$ . How much work would it take to move the object over #x in [2, 3], where x is in meters?

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An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 6 x - 1$ $u_k(x)= 6x-1$ . How much work would it take to move the object over #x in [2, 3], where x is in meters?

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Answer 1 · 2016-01-19T00:49:40

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Explanation:

The question is similar to another question solved as bellow. The difference is in the value of coefficient of friction. Choose appropriate value and solve.
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Force of kinetic friction which needs to be overcome to move the object

$F_{k} =$ $F_k=$ Coefficient of kinetic friction $μ_{k} \times$ $mu_ktimes$ normal force $η$ $eta$
where $η = m g$ $eta=mg$
Inserting given quantities and taking the value of $g = 9.8 m / s^{2}$ $g=9.8 m//s^2$
$F_{k} = (x + 3) \times 5 \times 9.8 N$ $F_k=(x+3)times 5times 9.8 N$
$F_{k} = 49 (x + 3) N$ $F_k=49(x+3) N$

When this force moves through a small distance $d x$ $dx$ , the work done is given as
$F_{k} d x = 49 (x + 3) d x$ $F_k dx=49(x+3) dx$
When the force moves through a distance from $x \in [2, 3]$ $x in [2, 3]$ , total work done is integral of RHS over the given interval.

Total work done $= \int_{2}^{3} 49 (x + 3) d x$ $=int_2^3 49(x+3)dx$
$\Rightarrow$ $implies$ Total work done $= 49 \int_{2}^{3} (x + 3) d x$ $=49 int_2^3 (x+3)dx$

$= 49 (\frac{x^{2}}{2} + 3 x + C) {∣ ∣ ∣}_{2}^{3}$ $=49 (x^2 /2+3x+C)|_2^3$ , where C is constant of integration.
$= 49 [(\frac{3^{2}}{2} + 3 \times 3 + C) - (\frac{2^{2}}{2} + 3 \times 2 + C)]$ $=49 [ (3^2 /2+3times3+C)-(2^2 /2+3times2+C)]$
$= 49 [\frac{9}{2} + 9 - 2 - 6]$ $=49 [ 9 /2+9-2-6]$
$= 49 [4.5 + 1]$ $=49 [ 4.5+1]$

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An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 6 x - 1$ $u_k(x)= 6x-1$ . How much work would it take to move the object over #x in [2, 3], where x is in meters?

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Explanation:

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An object with a mass of 5kg5 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=6x−1u_k(x)= 6x-1 . How much work would it take to move the object over #x in [2, 3], where x is in meters?

1 Answer

Explanation:

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An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 6 x - 1$ $u_k(x)= 6x-1$ . How much work would it take to move the object over #x in [2, 3], where x is in meters?