An object with a mass of $1 k g$ $1 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = e^{x} - x + 3$ $u_k(x)= e^x-x+3$ . How much work would it take to move the object over #x in [1, 4], where x is in meters?

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Answer 1 · 2018-03-14T16:25:14

The work is $= 523.1 J$ $=523.1J$

$Reminder :$ $"Reminder : "$

$\int e^{x} d x = e^{x} + C$ $inte^xdx=e^x+C$

The work done is

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

The frictional force is

$F_{r} = μ_{k} \cdot N$ $F_r=mu_k*N$

The coefficient of kinetic friction is $μ_{k} = (e^{x} - x + 3)$ $mu_k=(e^x-x+3)$

The normal force is $N = m g$ $N=mg$

The mass of the object is $m = 1 k g$ $m=1kg$

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

$= 1 \cdot (e^{x} - x + 3) g$ $=1*(e^x-x+3)g$

The work done is

$W = 1 g \int_{1}^{4} (e^{x} - x + 3) d x$ $W=1gint_(1)^(4)(e^x-x+3)dx$

$= 1 g \cdot {[e^{x} - \frac{x^{2}}{2} + 3 x]}_{1}^{x}$ $=1g*[e^x-x^2/2+3x]_(1)^(4)$

$= 1 g ((e^{4} - 8 + 12) - (e^{1} - \frac{1}{2} + 3))$ $=1g((e^4-8+12)-(e^1-1/2+3))$

$= 1 g (e^{4} - e + \frac{3}{2})$ $=1g(e^4-e+3/2)$

$= 523.1 J$ $=523.1J$

An object with a mass of 1kg1 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=ex−x+3u_k(x)= e^x-x+3 . How much work would it take to move the object over #x in [1, 4], where x is in meters?