An object with a mass of $6 k g$ $6 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + 5 cot x$ $u_k(x)= 1+5cotx$ . How much work would it take to move the object over $x \in [\frac{11 π}{12}, \frac{7 π}{8}]$ $x in [(11pi)/12, (7pi)/8]$ , where x is in meters?

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An object with a mass of $6 k g$ $6 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + 5 cot x$ $u_k(x)= 1+5cotx$ . How much work would it take to move the object over $x \in [\frac{11 π}{12}, \frac{7 π}{8}]$ $x in [(11pi)/12, (7pi)/8]$ , where x is in meters?

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Answer 1 · 2017-08-17T18:15:41

$W \approx 107 J$ $W~~107"J"$

Explanation:

Work done by a variable force is defined as the area under the force vs. displacement curve. Therefore, it can be expressed as the integral of force:

$W = \int_{x_{i}}^{x_{f}} F_{x} d x$ $color(darkblue)(W=int_(x_i)^(x_f)F_xdx)$

where $x_{i}$ $x_i$ is the object's initial position and $x_{f}$ $x_f$ is the object's final position

Assuming dynamic equilibrium, where we are applying enough force to move the object but not enough to cause a net force and therefore not enough to cause an acceleration, our parallel forces can be summed as:

$\sum F_{x} = F_{a} - f_{k} = 0$ $sumF_x=F_a-f_k=0$

Therefore we have that $F_{a} = f_{k}$ $F_a=f_k$

We also have a state of dynamic equilibrium between our perpendicular forces:

$\sum F_{y} = n - F_{g} = 0$ $sumF_y=n-F_g=0$

$\Rightarrow n = m g$ $=>n=mg$

We know that ${\to f}_{k} = μ_{k} \to n$ $vecf_k=mu_kvecn$ , so putting it all together, we have:

${\to f}_{k} = μ_{k} m g$ $vecf_k=mu_kmg$

$\Rightarrow W = \int_{x_{i}}^{x_{f}} μ_{k} m g d x$ $=>color(darkblue)(W=int_(x_i)^(x_f)mu_kmgdx)$

We have the following information:

$\mapsto m = 6 kg$ $|->"m"=6"kg"$
$\mapsto μ_{k} (x) = 1 + 5 cot (x)$ $|->mu_k(x)=1+5cot(x)$
$\mapsto x \in [\frac{11 π}{12}, \frac{7 π}{8}]$ $|->x in[(11pi)/12,(7pi)/8]$
$\mapsto g = 9.81 m / s^{2}$ $|->g=9.81"m"//"s"^2$

Returning to our integration, know the $m g$ $mg$ quantity, which we can treat as a constant and move outside the integral.

$W = m g \int_{x_{i}}^{x_{f}} μ_{k} d x$ $color(darkblue)(W=mgint_(x_i)^(x_f)mu_kdx)$

Substituting in our known values:

$\Rightarrow W = (6) (9.81) \int_{\frac{11 π}{12}}^{\frac{7 π}{8}} 1 + 5 cot (x) d x$ $=>W=(6)(9.81)int_((11pi)/12)^((7pi)/8)1+5cot(x)dx$

This is a basic integral.

$\Rightarrow W \approx 107.39$ $=>W~~107.39$

Therefore, we have that the work done is $\approx 107 J$ $~~107J$ .

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An object with a mass of 6kg6 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=1+5cotxu_k(x)= 1+5cotx . How much work would it take to move the object over x∈[11π12,7π8]x in [(11pi)/12, (7pi)/8], where x is in meters?

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