An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = x + x sin x$ $u_k(x)= x+xsinx$ . How much work would it take to move the object over $x \in [0, 8 π]$ $x in [0, 8pi]$ , 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 $u_{k} (x) = x + x sin x$ $u_k(x)= x+xsinx$ . How much work would it take to move the object over $x \in [0, 8 π]$ $x in [0, 8pi]$ , where x is in meters?

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Answer 1 · 2017-08-10T19:45:59

$W \approx 14259 J$ $W~~14259J$

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:

$\mapsto m = 5 kg$ $|->"m"=5"kg"$
$\mapsto μ_{k} (x) = x + x sin (x)$ $|->mu_k(x)=x+xsin(x)$
$\mapsto x \in [0, 8 π]$ $|->x in[0,8pi]$
$\mapsto g = 9.81 m / s^{2}$ $|->g=9.81"m"//"s"^2$

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:

$\Rightarrow W = (5) (9.81) \int_{0}^{8 π} (x + x sin x) d x$ $=>W=(5)(9.81)int_(0)^(8pi)(x+xsinx)dx$

Let's break the integral up using the sum rule.

$\Rightarrow = (5) (9.81) [\int_{0}^{8 π} (x) d x + \int_{0}^{8 π} (x sin x) d x]$ $=>=(5)(9.81)[int_(0)^(8pi)(x)dx+int_(0)^(8pi)(xsinx)dx]$

The left is a basic integral, but the right will require integration by parts to solve.

$u = x d u = d x$ $" "u=x " " du=dx$

$d v = sin (x) d x v = - cos (x)$ $dv=sin(x)dx " " v=-cos(x)$

We will have an integral of the form:

$u v - v \int d u$ $uv-vintdu$

Therefore, for the right most integral, we obtain:

$- x cos (x) - (\int_{0}^{8 π} (- cos x) d x)$ $-xcos(x)-(int_(0)^(8pi)(-cos x)dx)$

$\Rightarrow - x cos (x) + \int_{0}^{8 π} (cos x) d x$ $=>-xcos(x)+int_(0)^(8pi)(cosx)dx$

We now have a basic integral.

Putting it all together:

$(5) (9.81) [\int_{0}^{8 π} (x) d x + {- x cos (x) + \int_{0}^{8 π} (cos x) d x}]$ $(5)(9.81)[int_(0)^(8pi)(x)dx+{-xcos(x)+int_(0)^(8pi)(cosx)dx}]$

Which we can now evaluate, yielding $W = 14258.57$ $W=14258.57$

Therefore, we have that the work done is $\approx 14259 J$ $~~14259J$ or $1.4 \cdot 10^{4} J$ $1.4*10^4J$

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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 $u_{k} (x) = x + x sin x$ $u_k(x)= x+xsinx$ . How much work would it take to move the object over $x \in [0, 8 π]$ $x in [0, 8pi]$ , where x is in meters?

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Explanation:

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An object with a mass of 5 kg5kg5 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= x+xsinx uk(x)=x+xsinxu_k(x)= x+xsinx . How much work would it take to move the object over x in [0, 8pi]x∈[0,8π]x in [0, 8pi], where x is in meters?

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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 $u_{k} (x) = x + x sin x$ $u_k(x)= x+xsinx$ . How much work would it take to move the object over $x \in [0, 8 π]$ $x in [0, 8pi]$ , where x is in meters?