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Gelneren [198K]
3 years ago
15

You do 120 j of work while pulling your sister back on a swing, whose chain is 5.10 m long. you start with the swing hanging ver

tically and pull it until the chain makes an angle of 32.0° with the vertical with your sister is at rest. what is your sister's mass, assuming negligible friction?
Physics
1 answer:
Goryan [66]3 years ago
5 0
The work done to pull the sister back on the swing is equal to the increase in potential energy of the sister:
W= \Delta U = mg \Delta h (1)

where m is the sister's mass, g is the gravitational acceleration and \Delta h is the increase in altitude of the sister with respect to its initial position.

By calling \theta the angle of the chain with respect to the vertical, the increase in altitude is given by
\Delta h = L - L \cos \theta = L(1 - \cos \theta) (2)
where L is the length of the chain.

Putting (2) inside (1), we find
W= m g L (1 - \cos \theta)
from which we can find the mass of the sister:
m =  \frac{W}{g L (1 - \cos \theta)} =  \frac{120 J}{(9.81 m/s^2)(5.10 m)(1- \cos 32.0^{\circ})} =15.8 kg
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Answer:

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3 years ago
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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
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Explanation:

(a)  For an isothermal process, work done is represented as follows.

             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}})

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             = 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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Answer:

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