A rope is lying on the floor and has a mass which applies a force of 3 N/m. How much work is required to raise one end of the rope to a height of 2 meters?

In general, if it takes a force of [tex]F(h)[/tex] newtons to lift an object at height [tex]h[/tex], the work done lifting the object from [tex]h = a[/tex] to [tex]h = b[/tex] can be found using the definite integral about [tex]h[/tex]:

[tex]\displaystyle W = \int \limits_{a}^{b} F(h)\, dh[/tex].

If the value of [tex]F(h)[/tex] is a constant [tex]m \cdot g[/tex] regardless of height [tex]h[/tex], then the result of the integral would be

[tex]\displaystyle \int \limits_{a}^{b} (m \cdot g)\, dh = \left[m \, g\, h \right]_a^b = m\cdot g \, (b - a)[/tex].

However, in this case the value of [tex]F(h)[/tex] does depend on the value of [tex]h[/tex].

At height [tex]h = 0\; \rm m[/tex], nothing is being lifted. The amount of force required would be zero.
At height [tex]h = 1\; \rm m[/tex], one meter of the rope is in the air. That requires a force of at least [tex]1\; \rm m \times 3\; N \cdot m^{-1} = 3\; N[/tex].
In general, at a height of [tex]h[/tex] meters, the force required would be at least [tex]3\, h[/tex] Newtons.

In other words, [tex]F(h) = 3\; h[/tex] where [tex]F[/tex] is in Newtons and [tex]h[/tex] is in meters.

Evaluate the integral:

[tex]\begin{aligned} W &= \int \limits_{a}^{b} F(h)\, dh \cr &= \int \limits_{0}^{2}3\, h \, dh && \text{Apply the power rule.}\cr &= \left[\frac{3}{2}\,h^2\right]_{0}^{2}\cr &= \frac{3}{2} \times 2^2 \cr &= 6\end{aligned}[/tex].

whitneytr12 whitneytr12 · Answer 2 · 2021-09-09T15:58:25+02:00

The work required to raise one end of the rope to a height of 2 meters with an applied force of 3 N is 6 joules.

We can calculate the work required to raise one end of the rope to a height of 2 meters as follows:

[tex] W = F*d*cos(\theta) [/tex]

Where:

F: is the force exerted = 3 N (newton: unit of force)

d: is the displacement = 2 m

θ: is the angle between the applied force and the displacement

Since the force and the displacement are in the same direction, θ = 0, so:

[tex] W = F*d*cos(\theta) = Fdcos(0) = F*d [/tex]

Hence, the work done is:

[tex] W = 3 N*2m = 6 J [/tex]

Therefore, it is required 6 J of work.

You can see another example of work calculation here: https://brainly.com/question/2270290?referrer=searchResults

I hope it helps you!