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

Let $T$ be the tension in the chain and $N$ be the normal reaction.

Since the box is slid slowly, the net force on it is zero.

Balancing the forces in the vertical direction:

$N + T \sin\theta = mg$

$N = mg - T \sin\theta$

Balancing the forces in the horizontal direction:

$T \cos\theta = \mu N$

Substituting the value of $N$:

$T \cos\theta = \mu (mg - T \sin\theta)$

$T \cos\theta + \mu T \sin\theta = \mu mg$

$T \cos\theta (1 + \mu \tan\theta) = \mu mg$

$T \cos\theta = \dfrac{\mu mg}{1 + \mu \tan\theta}$

The work done by the person is the work done by the tension force over the distance $d$:

$W = (T \cos\theta) d$

Substituting $mg = 2000 \text{ N}$, $d = 20 \text{ m}$, and $\mu = 0.2$:

$W = \dfrac{0.2 \times 2000 \times 20}{1 + 0.2 \tan\theta}$

$W = \dfrac{8000}{1 + \dfrac{\tan\theta}{5}}$

$W = \dfrac{40000}{5 + \tan\theta} \text{ J}$