# An object with a mass of 5 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= x+xsinx . How much work would it take to move the object over #x in [0, 5pi], where x is in meters?

Mar 13, 2016

$7 \text{kJ}$

#### Explanation:

The weight of the object is given by $W = m \cdot g$. Since the object is not accelerating in the vertical direction, the normal force, $N$, is exactly equal to the weight.

The friction is given by $F = {u}_{\text{k}} \cdot N$.

Consider the object being dragged/pushed along a small distance $\text{d} x$. The amount of work needed is $F \cdot \text{d"x = N u_k(x) "d} x$.

The total work done to oppose the friction is given by

${\int}_{0}^{5 \pi} F \mathrm{dx} = N {\int}_{0}^{5 \pi} {u}_{k} \left(x\right) \text{d} x$

$= m g {\int}_{0}^{5 \pi} \left(x + x \sin \left(x\right)\right) \text{d} x q \quad$ (integrate by parts)

$= m g {\left[{x}^{2} / 2 - x \cos \left(x\right) + \sin \left(x\right)\right]}_{0}^{5 \pi}$

$= m g \left(\frac{25 {\pi}^{2}}{2} + 5 \pi\right)$

$= \left(5 {\text{kg") (9.81 "m/s}}^{2}\right) \left(\frac{25 {\pi}^{2}}{2} + 5 \pi\right)$

$= 6821 \text{J}$