An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = e^{x} - 2 x + 3$ $u_k(x)= e^x-2x+3$ . How much work would it take to move the object over #x in [3, 4], 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 $u_{k} (x) = e^{x} - 2 x + 3$ $u_k(x)= e^x-2x+3$ . How much work would it take to move the object over #x in [3, 4], where x is in meters?

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Answer 1 · 2017-08-05T17:50:00

$W = 1497$ $W = 1497$ $J$ $"J"$

Explanation:

We're asked to find the necessary work that needs to be done on a $5$ $5$ - $kg$ $"kg"$ object to move on the position interval $x \in [3 l m, 4 l m]$ $x in [3color(white)(l)"m", 4color(white)(l)"m"]$ with a varying coefficient of kinetic friction $μ_{k}$ $mu_k$ represented by the equation

$ul(mu_k(x) = e^x - 2x + 3$

The work $W$ done by the necessary force is given by

$ulbar(|stackrel(" ")(" "W = int_(x_1)^(x_2)F_xdx" ")|)$ $color(white)(a)$ (one dimension)

where

$F_x$ is the magnitude of the necessary force
$x_1$ is the original position ( $3$ $"m"$ )
$x_2$ is the final position ( $4$ $"m"$ )

The necessary force would need to be equal to (or greater than, but we're looking for the minimum value) the retarding friction force, so

$F_x = f_k = mu_kn$

Since the surface is horizontal, $n = mg$ , so

$ul(F_x = mu_kmg$

The quantity $mg$ equals

$mg = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 49.05color(white)(l)"N"$

And we also plug in the above coefficient of kinetic friction equation:

$ul(F_x = (49.05color(white)(l)"N")(e^x-2x+3)$

Work is the integral of force, and we're measuring it from $x=3$ $"m"$ to $x=4$ $"m"$ , so we can write

$color(red)(W) = int_(3color(white)(l)"m")^(4color(white)(l)"m")(49.05color(white)(l)"N")(e^x-2x+3) dx = color(red)(ulbar(|stackrel(" ")(" "1497color(white)(l)"J"" ")|)$

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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 $u_{k} (x) = e^{x} - 2 x + 3$ $u_k(x)= e^x-2x+3$ . How much work would it take to move the object over #x in [3, 4], 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)= e^x-2x+3 uk(x)=ex−2x+3u_k(x)= e^x-2x+3 . How much work would it take to move the object over #x in [3, 4], 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 $u_{k} (x) = e^{x} - 2 x + 3$ $u_k(x)= e^x-2x+3$ . How much work would it take to move the object over #x in [3, 4], where x is in meters?