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Vinvika [58]
2 years ago
15

A slinky is traveling down the stairs, like in the video clip below. What is the total KINETIC ENERGY of the slinky at the botto

m of the stairs (just before it stops moving) IF the Height of the stairs is 2 meters, the weight of the slinky is 4.41 Newtons, its spring constant is 0.84 N/m, and the distance the slinky is initially stretched (to get it going) is 0.25 meters??
Physics
1 answer:
yuradex [85]2 years ago
7 0

Answer:

8.79 J

Explanation:

Given that a slinky is traveling down the stairs, like in the video clip below. What is the total KINETIC ENERGY of the slinky at the bottom of the stairs (just before it stops moving) IF the Height of the stairs is 2 meters, the weight of the slinky is 4.41 Newtons, its spring constant is 0.84 N/m, and the distance the slinky is initially stretched (to get it going) is 0.25 meters??

Total energy = mgh

Total energy = 4.41 × 2 = 8.82 J

Elastic potential energy = 1/2 × Ke^2

Elastic potential energy = 1/2 × 0.84 × 0.25^2

Elastic potential energy = 0.02625

Also,

Total energy = P.E + K.E

Substitute them into the formula above

8.82 = 0.02625 + K.E

K.E = 8.82 - 0.02625

K.E = 8.79375

K.E = 8.79 J

Therefore, the KINETIC ENERGY of the slinky at the bottom of the stairs is 8.79 Joules approximately

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             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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