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dimaraw [331]
2 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]2 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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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
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y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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