An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 4 + sec x$ $u_k(x)= 4+secx$ . How much work would it take to move the object over #x in [0, (pi)/12], where x is in meters?

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Answer 1 · 2017-04-13T08:21:28

The work is $= 89.7 J$ $=89.7J$

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We need

$\int sec x d x = ln (tan x + sec x)$ $intsecxdx=ln(tanx+secx)$

The frictional force is

$F = F_{r} = μ_{k} \cdot N$ $F=F_r=mu_k*N$

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

So,

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

But,

$μ_{k} (x) = 4 + sec x$ $mu_k(x)=4+secx$

The work done is

$W = F \cdot d = \int_{0}^{\frac{π}{12}} μ_{k} m g d x$ $W=F*d=int_0^(pi/12)mu_kmgdx$

$= 7 g \int_{0}^{\frac{π}{12}} (4 + sec x) d x$ $=7gint_0^(pi/12)(4+secx)dx$

$= 7 g {[4 + ln (tan x + \frac{1}{cos x})]}_{0}^{\frac{π}{12}}$ $=7g[4+ln(tanx+1/cosx)]_0^(pi/12)$

$= 7 g ((\frac{π}{3} + 0.26) - (0))$ $=7g((pi/3+0.26)-(0))$

$= 7 g \cdot 1.31$ $=7g*1.31$

$= 89.7 J$ $=89.7J$

An object with a mass of 7kg7 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=4+secxu_k(x)= 4+secx . How much work would it take to move the object over #x in [0, (pi)/12], where x is in meters?