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timurjin [86]
3 years ago
5

Consider a system consisting of the Earth, a vertical spring attached to the Earth, and a mass attached to the other end of the

spring. Beginning with the mass hanging in its equilibrium position, a human grips the mass and pulls it down by some distance and holds the mass at rest. How does conservation of energy apply to this action?
Physics
2 answers:
motikmotik3 years ago
8 0

Answer:

The gravitational potential energy was not only converted to elastic potential energy, but the positive work the human did on the system also added to the increase in elastic potential energy.

Explanation:

Let's first of all choose our system to consist of the block, spring, and Earth.

By choosing the Earth and block to be in the system, we will have a change in gravitational potential energy.

Now, in the beginning, the spring hangs vertically with a block of mass (m) attached at the bottom.

During the pulling process by the human, he is doing positive work on the system and this means that the energy in the system increases.

Let's now use the work-energy theorem to answer the question of where the gravitational potential energy goes;

W_net,external = ΔE_total = -ΔU_gravity + ΔU_elastic

So, we see the gravitational potential energy decreases. However, the only other term that could compensate for this decrease in gravitational potential energy is the increase in elastic potential energy.

The spring did not store gravitational potential energy but instead, the gravitational potential energy was converted to elastic potential energy.

From the work energy theorem equation above, since the left-hand side is positive, the absolute value of ΔU_elastic would be greater than that of ΔU_gravity.

Therefore, we see that the gravitational potential energy was not only converted to elastic potential energy, but the positive work the human did on the system also adds to the increase in elastic potential energy.

Liula [17]3 years ago
7 0

Answer:

The external force leads to an increase on gravitational and spring potential energies.

Explanation:

The system consists of a mass, resort and Earth. According to the Principle of Energy Conservation there is a potential energy as a consequence of the interaction between Earth and the mass and spring potential energy because of the spring deformation and, besides, the existence of work due to an external force:

U_{g,A} + U_{k,A} + W_{ext} = U_{g,B} + U_{k,B}

W_{ext} = (U_{g,B}-U_{g,A}) + (U_{k,B}-U_{k,A})

F\cdot \Delta r = \frac{G\cdot m\cdot M}{r_{o}-\Delta r} - \frac{G\cdot m \cdot M}{r_{o}} +\frac{1}{2}\cdot k \cdot (x_{o}+\Delta r)^{2} - \frac{1}{2} \cdot k \cdot (x_{o})^{2}

F\cdot \Delta r = G\cdot m \cdot M \cdot \left(\frac{1}{r_{o}-\Delta r}-\frac{1}{r_{o}}   \right)+\frac{1}{2}\cdot k \cdot [(x_{o}+\Delta r)^{2} -x_{o}^{2}]

The external force leads to an increase on gravitational and spring potential energies.

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What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
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<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

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