An object with a mass of $4 k g$ $4 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 5 + tan x$ $u_k(x)= 5+tanx$ . How much work would it take to move the object over #x in [(pi)/12, (5pi)/12], where x is in meters?

Question

An object with a mass of $4 k g$ $4 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 5 + tan x$ $u_k(x)= 5+tanx$ . How much work would it take to move the object over #x in [(pi)/12, (5pi)/12], where x is in meters?

1 Answer

Impact of this question

1384 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-06T06:20:23

The work is $= 257 J$ $=257J$

Explanation:

The work done is

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

The frictional force is

$F_{r} = μ_{k} \cdot N$ $F_r=mu_k*N$

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

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

$= 4 (5 + tan x)) g$ $=4(5+tanx))g$

The work done is

$W = 4 g \int_{\frac{1}{12} π}^{\frac{5}{12} π} (5 + tan x) d x$ $W=4gint_(1/12pi)^(5/12pi)(5+tanx)dx$

$= 4 g \cdot [5 x - ln | cos x |)]_{\frac{1}{12} π}^{\frac{5}{12} π}$ $=4g*[5x-ln|cosx|)]_(1/12pi)^(5/12pi)$

$= 4 g ((\frac{25}{12} π - ln ∣ ∣ ∣ cos (\frac{5}{12} π) ∣ ∣ ∣) - (\frac{5}{12} π - ln ∣ ∣ ∣ cos (\frac{1}{12} π) ∣ ∣ ∣))$ $=4g((25/12pi-ln|cos(5/12pi)|)-(5/12pi-ln|cos(1/12pi)|))$

$= 4 g (\frac{5}{3} π - ln ∣ ∣ ∣ cos (\frac{5}{12} π) ∣ ∣ ∣ + ln (cos (\frac{1}{12} π))$ $=4g(5/3pi-ln|cos(5/12pi)|+ln(cos(1/12pi))$

$= 4 g (\frac{5}{3} π + 1.32))$ $=4g(5/3pi+1.32))$

$= 4 g \cdot 6.56 J$ $=4g*6.56J$

$= 257 J$ $=257J$

Science

Math

Humanities

... and beyond

An object with a mass of $4 k g$ $4 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 5 + tan x$ $u_k(x)= 5+tanx$ . How much work would it take to move the object over #x in [(pi)/12, (5pi)/12], where x is in meters?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

An object with a mass of 4 kg4kg4 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 5+tanx uk(x)=5+tanxu_k(x)= 5+tanx . How much work would it take to move the object over #x in [(pi)/12, (5pi)/12], where x is in meters?

1 Answer

Explanation:

Related questions

Impact of this question

An object with a mass of $4 k g$ $4 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 5 + tan x$ $u_k(x)= 5+tanx$ . How much work would it take to move the object over #x in [(pi)/12, (5pi)/12], where x is in meters?