An object with a mass of $3 k g$ $3 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 3 x^{2} - 2 x + 12$ $u_k(x)= 3x^2-2x+12$ . How much work would it take to move the object over #x in [1, 3], where x is in meters?

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An object with a mass of $3 k g$ $3 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 3 x^{2} - 2 x + 12$ $u_k(x)= 3x^2-2x+12$ . How much work would it take to move the object over #x in [1, 3], where x is in meters?

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Answer 1 · 2017-08-05T23:03:09

$W = 1236$ $W = 1236$ $J$ $"J"$

Explanation:

We're asked to find the necessary work that needs to be done on a $3$ $3$ - $kg$ $"kg"$ object to move on the position interval $x \in [1 l m, 3 l m]$ $x in [1color(white)(l)"m", 3color(white)(l)"m"]$ with a varying coefficient of kinetic friction $μ_{k}$ $mu_k$ represented by the equation

$ul(mu_k(x) = 3x^2-2x+12$

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 ( $1$ $"m"$ )
$x_2$ is the final position ( $3$ $"m"$ )

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

$mg = (3color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 29.43color(white)(l)"N"$

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

$ul(F_x = (29.43color(white)(l)"N")(3x^2-2x+12)$

Work is the integral of force, and we're measuring it from $x=1$ $"m"$ to $x=3$ $"m"$ , so we can write

$color(red)(W) = int_(1color(white)(l)"m")^(3color(white)(l)"m")(29.43color(white)(l)"N")(3x^2-2x+12) dx = color(red)(ulbar(|stackrel(" ")(" "1236color(white)(l)"J"" ")|)$

Answer 2 · 2017-08-05T23:39:33

We observe that Coefficient of kinetic friction is given as

$u_k(x)=3x^2−2x+12$

and that the object is being pushed along a linear path.

We know that force of friction opposes the motion. As such we can infer that movement is only in the $x$ -direction.
The angle between force of friction and displacement is $180^@$

Work Done against force $vecF$ for moving a distance $vecs$ is given as

$W=vecFcdotvecs$

Let the object move a distanace $vecdx$ . Work done to move the object against the force of friction

$dW=vec(F_f)cdot vecdx$

Force of friction $vec(F_f)=-u_kNhatx$
where $N$ is normal reaction and given $=mg$
$dW=-(3x^2−2x+12)mghatxcdot vecdx$

Integrating both sides for the given interval we get
$W=-int_1^3(3x^2−2x+12)mgdx$

$W=-3xx9.81|3x^3/3−2x^2/2+12x|_1^3$
$W=-29.43[(3^3−3^2+12xx3)-(1^3−1^2+12xx1)]$
$W=-29.43[(27−9+36)-(1−1+12)]$
$W=-29.43(54-12)$
$W=-1236J$

$-ve$ sign shows that this amount of energy was lost in doing work against friction.

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An object with a mass of $3 k g$ $3 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 3 x^{2} - 2 x + 12$ $u_k(x)= 3x^2-2x+12$ . How much work would it take to move the object over #x in [1, 3], where x is in meters?

2 Answers

Explanation:

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An object with a mass of 3 kg3kg3 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 3x^2-2x+12 uk(x)=3x2−2x+12u_k(x)= 3x^2-2x+12 . How much work would it take to move the object over #x in [1, 3], where x is in meters?

2 Answers

Explanation:

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An object with a mass of $3 k g$ $3 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 3 x^{2} - 2 x + 12$ $u_k(x)= 3x^2-2x+12$ . How much work would it take to move the object over #x in [1, 3], where x is in meters?