Question

A 3 meter chain with linear mass density rho(x) = 2x(4 − x) kg/m lies on the ground, where x = 0 is the top of the chain. Calculate the work required to lift the chain from its top end so that its bottom is 2 meters above the ground.

Answer 1

the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J

Explanation:

Given that:

The mass density of the chain is
`\rho (x) = 2x( 4-x) \ kg/m`

It is pertinent and crucial to consider the determination of the work-done in lifting the chain from its front in that its bottom is 2 meters away from the ground.

Consider a cross section portion of the chain of length Δx that has to be lifted to a height
`x_k`

The required wok to be done for this work is
`W_k = \rho(x_k)g \delta x(x_k)`

combining the segments of the chain and taking the
`\delta x`→0

Integrally, the work-done can be illustrated as :

`\int\limits^3_0 \rho {(x_k)} \, gxdx \ \ = \ \ \int\limits^3_0 (9.8) 2x^2 (4-x)dx \\\\= 19.6 \int\limits^3_0 (4x^2-x^3)dx\\\\\\= 19.6 [ (4)/(3)x^2- \frac {x^4}{4}]^3__0}}dx\\\\\\= 19.6 [ (4)/(3)(3)^2- \frac {3^4}{4}]}dx\\\\= 19.6 (36- (81)/(4))\\\\\\= 308.7 \ \ J`

Furthermore, there is need to lift the chain up to 2 meters . So, calculating the weight of the chain ; we have:

Weight =
`\int\limits^3_0 \rho {x} \, gdx`

`= \int\limits^3_0 (9.8) {2x}(4-x) \, dx\\ \\\\= 19.6\int\limits^3_0 {(4x-x^2)} \, dx \\\\= 19.6 [ 2x^2 - (x^3)/(3)]^3_0\\\\= 19.6 [2(3)^2 - (3^3)/(3)]\\\\=19.6 [18-9]\\\\= 176.4 \ \ J`

Finally .the work-done is said to be equal to the potential energy

∴ W = mgh

W = (176.4)×2

W = 352.8 J

Total workdone = (308.7 + 352.8 ) J

Total workdone = 661 J

Thus, the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J

Answer 2

Explanation:

Answer:

Explanation:

Given that,

Length of chain

L = 3m

Linear mass density is

ρ(x) = 2x(4 — x) kg/m lies on the ground

When x = 0, is top of the chain

Work done to lift the chain from top end so that the bottom is 2m above the ground.

Considered the segment of the chain of length ∆x that will be lifted in the positive y direction (+j)m from the foot.

The work needed to lift this segment is given as

Work = mass density × ∆x × gravity

W = ρ × ∆x × g

g is acting downward = 9.8j

Summing over all segment of the chain and passing to the limit as ∆x→0.

Therefore, the total work done needed to full extend the chain is

W = ∫ ρ × ∆x × g x = 0 to 3

Since g is constant

∆x = xdx

Then,

W = g ∫ 2x(4—x)x dx. x= 0 to x = 3

W = 9.81 ∫ (8x² — 2x³)x dx

W = 9.81 ( 8x³/3 — 2x⁴/4)

W=9.81(8x³/3— ½x⁴) from x=0 to x=3

W = 9.81[8(3)³/3 — ½(3⁴)] — 0

W = 9.81 × (72 —40.5)

W = 9.81 × 31.5

Work done= 309.015 J

Lifting the entire chain requires to light the weight

Weight = ∫ρgdx

Weight = g ∫ρ dx. From x=0 to x=3

Weight = g ∫ 2x(4-x) dx

Weight = 9.81 ∫(8x-2x²)dx

Weight = 9.81 [ 8x²/2 - 2x³/3]

Weight = 9.81[4x²-⅔x³] x=0 to x=3

Weight = 9.81[4(3²) — ⅔(3³)]

Weight = 9.81(36—18)

Weight = 9.81 × 18

Weight = 176.58N

Now, this weight is lifted to a height of 2m, then using potential energy formula, we have

P.E = Work = mgh = Weight ×height

Work = W×h = 176.58 × 2

Work = 353.16 J

Then, total workdone is

W = 353.16 + 309.015

W = 662.18 J

The required Work done required to lift the chain from top so that it's bottom is 2m from the ground is 662.18J