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## Homework Statement

(direct translations from Finnish to English to the best of my ability)

Ascertain whether the statements are correct or false. Give reasoning behind all answers.Correct any false statements. (direct translations from Finnish to English)

a)the change of an object's potential energy doubles during a lift, when the lifting height doubles.

b)When the speed of a falling object is half of its final speed, the potential energy of that object has been reduced to half of the original.

## Homework Equations

m*g*h= E

_{pot.}potential energy

1/2 * m * v

^{2}= E

_{kin.}kinetic energy

d= 1/2 * a * t

^{2}

delta t = (delta v) / a

## The Attempt at a Solution

I just wanted to doublecheck my results, because the physicsbook does not provide answers.

a) To my eyes it looks like the statement is true. Unless there's something very fishy about the wording of the question.

I dabbled with the equation mgh=E

I assumed m=1kg , g=9,81m/s

^{2}, h=1m

Obviously it seems that we must keep the mass and acceleration as constants. We affix the mass, and the acceleration to constant values, and we only change the height value. We compare doubled-height-scenario, with the original scenario. It's probably not the most rigorous proof that's out there, but you could make a chart, of the values, and compute the end result. Then you compare the potential energies to each other.

assuming that: g=9,81, m=1kg, h=1m = 9,81 Joules = scenario 1

9,81m/s

^{2}* 1kg * (2m)= 19,62 Joules= scenario 2

we can see that the increased amount of energy is +100%. This means the energy becomes doubled, as the height doubles, whereas the mass and acceleration due to gravity remain the same. The amount of increase, always is the same +100%, when the height doubles. So, whether the original statement was true or false seems debatable.

b)

to my eyes it looks like the statement is false.

However it took some time for me to use those equations to give a result. (especially this one : d= 0,5 * a * t

^{2})

I suppose one could say that when the object is held stationary at height of 2m. Then the object has zero kinetic energy and 100% of its supposed potential energy.

When falling object has reached its maximum velocity just before impact at ground. Then that objects kinetic energy is 100% of its supposed kinetic energy. However, it seems that the potential energy is at minimum of zero. When the object is lying on the floor, The height difference between the object and the floor is pretty much zero. Using the zero-product property from mathematics then the potential energy ought to be zero at least regarding the comparison of elevation between the object and the floor, when the object is directly upon the floor.

The question especially noted that the falling object's velocity should be half of the "endgame velocity". Original velocity at the beginning is assumed as zero by myself.

I guess we could calculate kinetic energy such that velocity is (end.velocity)/2

we assume h=2m, m=1kg, a=9,81m/s

^{2}E

_{pot}= 19,62 joules

we assume that energy is transformed from potential energy into kinetic energy at 100% effectiveness.

we could then calculate the "endgame-velocity" of falling object

19,62= 0,5 * 1kg * v

^{2}

v

^{2}= 19,62/0,5

v=6,2641839053 m/s

But we were tasked with 0,5 * endgame-velocity = (3,1320919527 m/s)

At this point we could I suppose, find the kinetic energy, with that halved speed, and the same mass (1kg)

0,5 * 1kg * 9,81m

^{2}/s

^{2}

=E

_{kin.}= 4,9050 Joules

If we compare 4,9050 joules with 19,62 joules. We can see that (4,9050)/ (19,62) = 1/4 = 25%

This tells us that the kinetic energy has increased from 0% to 25% of the original kinetic energy. I suppose, that the rest of the energy must be 75%. 75%+25% = 100%. Potential energy is 75% of the original energy

(again assuming 100% energy transformation efficiency from potential to kinetic and between)

0,75 * 19,62 = 14,715 joules

I suppose one could double-check that result thusly. It took some effort to find especially one of the equations . ( this one; d= 0,5 * a * t

^{2})

d t= d v / a

this gives

t-0 = (3,1320m/s -0) / 9,81m/s

^{2}

t= 0,3192 seconds

d= 0,5 * 9,81m/s

^{2}* (0,3192 s)

^{2}

ca. 0,5 metres distance travelled. Object travels this amount downwards from the original drop height.

2-0,5m = 1,5m

Epot= 1kg * 9,81m/s

^{2}* 1,5m= 14,715 joules

14,715 joules / 19,62 joules = 0,75 = 75%