Answer:
Approximately [tex]4.70 \times 10^{2}\; \rm J[/tex], assuming that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].
Step-by-step explanation:
When the force [tex]F[/tex] is constant, moving by [tex]\Delta x[/tex] in the direction of the force would do a work of [tex]F \cdot \Delta x[/tex].
However, in this example, as an increasing portion of the chain gets lifted up, the force lifting the chain would increase in size. Hence, apply the integral form of this equation.
Let [tex]F(x)[/tex] denote the force required to lift the chain when the height above the ground is [tex]x[/tex] meters. At that moment, exactly [tex](x / 10)[/tex] of the chain would be above the ground. The mass of that much of the chain would be [tex](x / 10)\![/tex] the mass of the whole chain.
[tex]\begin{aligned}F(x) &= m(x) \cdot g \\ &= 60 \cdot \frac{x}{10} \cdot g = 6\, g\, x\end{aligned}[/tex].
Integrate with respect to [tex]x[/tex]:
[tex]\begin{aligned}\int \limits_{0}^{4} F(x) \cdot {\rm d} x &= \int \limits_{0}^{4} 6\, g\, x \cdot {\rm d} x \\ &= \left[(3\, g)\, x^{2}\right]_{x=0}^{x=4}\\ &= 48\, g \approx 4.7 \times 10^{2} \end{aligned}[/tex].
In other words, it would take approximately [tex]4.7 \times 10^{2}\; \rm J[/tex] of energy to raise one end of the chain to that height of [tex]4\; \rm m[/tex].
