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

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Answer 1 · 2017-05-07T06:56:43

283.53J

Explanation:

First we will write the information provided
$μ_{k} = e^{x} + x$ $mu_k=e^x+x$
M= $5 K g$ $5 Kg$

we know that the work done by us to the 5 kg mass is
$W = F_{a p p l i e d} \cdot d s$ $W=F_(applied)*ds$

but in this case there is friction is involved which can be included in the equation by ,

$W = (F_{a p p l i e d} - f_{k}) \cdot d S$ $W=(F_(applied)-f_(k))*dS$

here we need to move the block without acceleration so the applied force must be equal to the kinetic friction .
$F_{a p p l i e d} = f_{k}$ $F_(applied)=f_k$

similarly the work done will also be the same in both the cases so the work done by the frictional force is given by .

$W = f_{k} \cdot d x = μ_{k} \cdot N \cdot d x = \int_{1}^{2} (e^{x} + x) \cdot m \cdot g \cdot d x = 283.53 J$ $W=f_k*dx=mu_k*N*dx=int_1^2(e^x+x)*m*g*dx=283.53J$

Answer 2 · 2017-05-07T07:01:09

The work is $= 302.3 J$ $=302.3J$

Explanation:

The work done is

$W = F \cdot d$ $W=F*d$

The frictional force is

$F_{r} = μ_{k} \cdot N$ $F_r=mu_k*N$

$N = m g$ $N=mg$

$F_{r} = μ_{k} \cdot m g$ $F_r=mu_k*mg$

$= 5 (e^{x} + x) g$ $=5(e^x+x)g$

The work done is

$W = 5 g \int_{1}^{2} (e^{x} + x) d x$ $W=5gint_(1)^(2)(e^x+x)dx$

$= 5 g \cdot {[e^{x} + \frac{x^{2}}{2}]}_{1}^{x}$ $=5g*[e^x+x^2/2]_(1)^(2)$

$= 5 g ((e^{2} + 2) - (e + \frac{1}{2})))$ $=5g((e^2+2)-(e+1/2)))$

$= 5 g (9.39 - 3.22)$ $=5g(9.39-3.22)$

$= 5 g \cdot 6.17$ $=5g*6.17$

$= 302.3 J$ $=302.3J$

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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 $u_{k} (x) = e^{x} + x$ $u_k(x)= e^x+x$ . How much work would it take to move the object over #x in [1, 2], where x is in meters?

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Explanation:

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)=ex+xu_k(x)= e^x+x . How much work would it take to move the object over #x in [1, 2], where x is in meters?

2 Answers

Explanation:

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 $u_{k} (x) = e^{x} + x$ $u_k(x)= e^x+x$ . How much work would it take to move the object over #x in [1, 2], where x is in meters?