A box weighing 2000 N is to be slowly slid a distance of 20 m along...

Let $F$ be the pulling force applied at an angle $\theta$ with the horizontal. For the box to slide slowly, the net force on it must be zero.

Resolving forces horizontally and vertically:

$F \cos\theta = \mu N$

$N + F \sin\theta = W$

Substituting $N = W - F \sin\theta$ into the first equation:

$F \cos\theta = \mu (W - F \sin\theta)$

$F (\cos\theta + \mu \sin\theta) = \mu W$

$F = \dfrac{\mu W}{\cos\theta + \mu \sin\theta}$

To minimize the pulling force $F$, the denominator $\cos\theta + \mu \sin\theta$ must be maximized. Taking the derivative with respect to $\theta$ and equating to zero:

$-\sin\theta + \mu \cos\theta = 0 \Rightarrow \tan\theta = \mu$

For $\tan\theta = \mu$, we have $\cos\theta = \dfrac{1}{\sqrt{1+\mu^2}}$ and $\sin\theta = \dfrac{\mu}{\sqrt{1+\mu^2}}$.

The minimum force is:

F_\min = \dfrac{\mu W}{\dfrac{1}{\sqrt{1+\mu^2}} + \dfrac{\mu^2}{\sqrt{1+\mu^2}}} = \dfrac{\mu W}{\sqrt{1+\mu^2}}

The work done by the person is the work done by the pulling force:

W_{done} = F_\min d \cos\theta

$W_{done} = \left(\dfrac{\mu W}{\sqrt{1+\mu^2}}\right) d \left(\dfrac{1}{\sqrt{1+\mu^2}}\right) = \dfrac{\mu W d}{1+\mu^2}$

Substituting the given values $\mu = 0.2$, $W = 2000 \text{ N}$, and $d = 20 \text{ m}$:

$W_{done} = \dfrac{0.2 \times 2000 \times 20}{1 + (0.2)^2} = \dfrac{8000}{1.04}$

$W_{done} = \dfrac{8000}{1.04} \approx 7692 \text{ J}$