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dimaraw [331]
3 years ago
7

A 60​-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

chain has a density of 10 ​kg/m. Use 9.8 m divided by s squared for the acceleration due to gravity. a. How much work is required to wind the entire chain onto the cylinder using the​ winch? b. How much work is required to wind the chain onto the cylinder if a 35​-kg block is attached to the end of the​ chain?
Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

  • Density of the chain = \rho\ =\ 10\ kg/m.
  • Length of the chain = L = 60 m
  • Acceleration due to gravity = g = 9.81 m/s^2

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = \rho dy

Work done due to the small element,

dw\ =\ \rho g (60\ -\ y)dy\\

Total work done to wind the entire chain = w

w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow  w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J

part (b)

  • mass of the block connected to the chain = m = 35 kg

Total work done to wind the chain = work done due to the chain + work done due to the mass

\therefore W\ =\ w\ +\ mgL\\\Rightarrow W\ =\ 176580\ +\ 35\times 9.81\times 60\\\Rightarrow W\ =\ 176580\ +\ 20601\\\Rightarrow W\ =\ 197381\ J

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Answer:

  1. a change in color
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Explanation:

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Read 2 more answers
3. A rocket is launched at an angle of 53 degrees above the 1 point
irina1246 [14]

Answer:

24,000 m

Explanation:

First find the rocket's final position and velocity during the first phase in the y direction.

Given:

v₀ = 75 sin 53° m/s

t = 25 s

a = 25 sin 53° m/s²

Find: Δy and v

Δy = v₀ t + ½ at²

Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²

Δy = 7736.8 m

v = at + v₀

v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)

v = 559.0 m/s

Next, find the final position of the rocket during the second phase (as a projectile).

Given:

v₀ = 559.0 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 15945.5 m

The total displacement is:

7736.8 m + 15945.5 m

23682.2 m

Rounded to two significant figures, the maximum altitude reached is 24,000 m.

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3 years ago
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