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siniylev [52]
3 years ago
10

A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate

what its kinetic energy will be just before hitting the ground.
Physics
1 answer:
babymother [125]3 years ago
6 0

Answer:

200 J

Explanation:

In this problem, I assume there is no air resistance, so the  system is isolated (=no external forces).

For an isolated system, the total mechanical energy is constant, and it is given by:

E=KE+PE

where

KE is the kinetic energy

PE is the potential energy

The kinetic energy is the energy due to the motion of the object,  while the potential energy is the energy due to the position of the object relative to the ground.

At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its  speed is zero; it means that all its mechanical energy is just potential energy, and it is:

E=PE_{max}=200 J

As the boulder falls  down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.

Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have

E=KE_{max}=200 J

Therefore, the kinetic energy just before hitting the ground is 200 J.

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Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

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Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

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Answer:

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