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

Question

1406 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-09T13:42:13

The work is $= 897.1 J$ $=897.1J$

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

The mass is $m = 3 k g$ $m=3kg$

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

$= 3 (e^{x} - 2 x + 3) g$ $=3(e^x-2x+3)g$

The work done is

$W = 3 g \int_{3}^{4} (e^{x} - 2 x + 3) d x$ $W=3gint_(3)^(4)(e^x-2x+3)dx$

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

$= 3 g ((e^{4} - 16 + 12) - (e^{3} - 9 + 9))$ $=3g((e^4-16+12)-(e^3-9+9))$

$= 3 g (e^{4} - e^{3} - 4)$ $=3g(e^4-e^3-4)$

$= 3 g (30.51)$ $=3g(30.51)$

$= 897.1 J$ $=897.1J$

An object with a mass of 3kg3 kg is pushed along a linear path with a kinetic friction coefficient of 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?