Question 13 and 14 on Motion

[taiga]

which school?

Essendon Grammar

I wouldn't say that

It was definitely 50.

I hope we get consequential though, I put 100 as well.

[m@tty]

100 comes from this which i posted in another thread:

I equated the energy at the top unstretched with the energy .4m below, the spring energy.

This gave:

mgh=(1/2)*k*x^2

which came out to k=100

As there is no loss in energy (as the hand does not take away from either sides energy)

8=8 Therefore there is a 0 change in energy, and it is conserved.

From a brief discussion we had with our teacher he agreed with us about this answer.

Think about the basics of the situation:
-Hand is lowering the spring gently => It is applying an upwards force on it => It takes energy to do this; work => Energy is not maintained in the spring-mass system.

[Davoo!]

Uh, what if you took gravity to 9.8 and got answers of 49 and 3.9J? No marks lost yes?

[Akirus]

Uh, what if you took gravity to 9.8 and got answers of 49 and 3.9J? No marks lost yes?

Yep, that's fine.

[axola]

100 comes from this which i posted in another thread:

I equated the energy at the top unstretched with the energy .4m below, the spring energy.

This gave:

mgh=(1/2)*k*x^2

which came out to k=100

As there is no loss in energy (as the hand does not take away from either sides energy)

8=8 Therefore there is a 0 change in energy, and it is conserved.

From a brief discussion we had with our teacher he agreed with us about this answer.

sorry mate, you cannot equate mgh to 0.5kx^2 due to the work done by the hand. If the ball was released, kinetic energy would remain in the sytem as well.

[schnappy]

Ok. Still with this are we? The hand did not do work. The hand 'slowly allowed the mass to fall' or something like that. This means the spring is being extended by the weight of the mass, the hand is not doing work.

It is lowered slowly so that it doesn't drop suddenly and oscillate.

[shpongle]

The hand did do work.
The weight of the mass also did work to the hand. At the beginning, the weight of the mass was held up completely by the hand (none by the spring), and as it began to be slowly lowered, more work was being done to the hand than the spring.
Gradually, more work was done to the spring than the hand so that eventually all of the weight force was applied to the spring.

However, this means that not all of the work done by the weight (gravitational potential energy) was converted into strain potential energy. Half of the gravitational potential energy went to strain, half went to the hand, which is why k was 50N/m not 100.

I said 100 in the exam

[_henwee]

Wasn't this a very similar question to Q5 in VCAA 2009 exam?
Except in this case... we equate gravitational potential energy to elastic, not kinetic to elastic.

[shpongle]

Gravitational potential energy wasn't equated to elastic energy though.
It said the hand 'gently' lowered the block of mass, which would mean that the block was always exerting some weight force onto the hand until it was finally released.
Gravitational potential energy is effectively the work of weight, and the work of the weight was not done completely to the spring, so the energies could not equate.

[karar]

  :DEinstein's Special relativity questions pleaaseeee

[a.zirek]

Okay look, if you had that spring and SUDDENLY moved your hand from underneath it, what would logically happen?

The spring would, obviously, extend more than 0.4m and then retract. And it would continue on oscillating until it reached its equilibrium point (which is in this case 0.4m).
That is why for this question you had to use F=kx, because the system is at equilibrium. The ONLY time you use E=mgh=o.5kx^2 is if the question says the mass FIRST came to rest at ...

The answer is most definitely 50Nm.

P.S. Sorry about the capital letters... just wanted to make sure people understand.

[ryansecond5]

The hand definately did work on the system. It exerted a force on the mass through a distance, thus performing work. In this case the work would be negative because the force and displacement vectors are in opposite directions.

[ddaly]

My teacher is the chief assessor!!!!
Question 13 - 50J

F=Kx    - Must use this formula as it was gently lowered

Question 14 - 4J      -Change in gravitational potential to spring potential and calculate how much is lost.

[Akirus]

The fact that the force of gravity is causing an extension of 0.40m is already enough to conclude that energy was lost.

[ddaly]

Exactly