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Artemon [7]
4 years ago
14

Two children of about the same weight are playing at the playground. They both climb up to the top of a small tower. One slides

back down to ground level on a frictionless slide. The other freely drops down from the tower. Both children start their trips back down to the ground with an initial speed of zero.
Which combination of statements below is most accurate? [Enter your answer as a string of the letter options you believe are correct. For instance, if you think options A, D, and E are correct, then enter "ADE"]

A) The work done by gravity on each child is positive as they climb the tower.
B) The children have different kinetic energies on reaching ground level
C) The work done by gravity on both children is negative as they descend from the tower.
D) The overall work done by gravity is zero.
E) None of the above
Physics
1 answer:
olya-2409 [2.1K]4 years ago
4 0

Answer:

D

Explanation:

D) The overall work done by gravity is zero  

This statement is correct .

If m be the mass of each of the children and h be the height of tower

work done by gravity on the boys in going up = - mgh

it is so because force applied by gravity = mg downwards and displacement

is upwards

work done will be negative = - mgh

Work done by gravity on boys when they come down = + mgh because both force and displacement are downwards .

Hence total work done = - mgh + mgh = 0.

The children will have same kinetic energy as the inclined surface is friction-less so no energy will be dissipated hence addition of energy to boys in both the cases will be same.

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Answer:

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The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

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m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

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