# Ek 1001 spring problem 248

Discussion in 'MCAT Study Question Q&A' started by TheMightyTexan, Sep 7, 2014.

1. ### TheMightyTexan 2+ Year Member

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The problem is simple, a spring is between a wall and a block on a friction-less surface, what happens after the mass is released BUT before the spring reaches the relaxed position.

I chose the velocity and the acceleration of the mass will continually decrease.

The answer is F=-kx, acceleration is in the difection of motion, increasing velocity. since deltaX decreases the force will be smaller resulting in decreased acceleration.

I'm calling BS on this question in general. It sounds like they chose (average) velocity and (instant) acceleration as the correct answer, leaving the other answers invalid. Since there are no stipulations on this matter, how is it that one is correct and the other not?

2. ### Cawolf 5+ Year Member

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Well we are looking to see what happens from this x = 5cm position to just before x = 0.

This means to me that the force is maximal at t=0 and will continually decrease to almost 0.

The acceleration is simple from looking at the force since a = f/m. The force will approach 0, so the acceleration decreases with time.

Then we look at the velocity. This block is being constantly positively accelerated to a maximal velocity at x = 0, so the velocity will constantly increase.

This corresponds with the book answer.

I think your error was not examining the velocity carefully enough - it is max at x = 0 and will not decrease until after it passes that point.

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### TheMightyTexan 2+ Year Member

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Ok i see where i mistook this problem..

I made an error in thinking too intuitively (friction always exists), and less accurately by not reading about the frictionless surface.
Now i see that velocity could not decrease, but it would not CONTINUOUSLY increase either, except at a small fraction of the distance.

Say the spring starts out accelerating the block at 5 m/s right at the 5 cm point, and decelerates at 1 m/s so that when the spring is fully extended at o cm, acceleration = 0. The box at will continuously increase in velocity until it reaches 5 m/s, and out run the spring leaving it behind and keeping it's 5 m/s velocity to infinity. So it would actually reach its maximal velocity soon before x = 0. If was connected to the spring, the same concept would result as if there was friction since the spring would slow it down.

All in all, a technical answer for a vague or misleading question. If i had thought of the friction less surface i believe i would have been able to pick the lesser evil. Kinda hate how the explanations in the book are so vague as well, but it makes sense they need to keep it concise.

Hopefully that all makes sense its been a long day..

Thanks @Cawolf for your help. I think ill just have to abandon ship when i encounter stuff like this since its so time absorbing and my test is in a week :/

4. ### mehc012 Big Damn Hero 5+ Year Member

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The most important point to take away here, which I feel you are missing, is that in an ideal spring system (or any similar example of harmonic motion), velocity is always maximum at zero displacement, and acceleration is always zero. This is not vague, nor specific to this setup. It is why SHM works. Most conceptual questions on the MCAT concerning this topic will come back to those basics, and very much resemble this question. Starting at max displacement, acceleration continually decreases and velocity continually increases until it reaches the equilibrium displacement (relaxed state), at which point acceleration reverses direction and velocity begins to decrease. When it reaches maximum displacement on the opposite side, velocity will be zero and acceleration will be maximal (but reversed from the initial value). Then those steps repeat with reversed signs.

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### TheMightyTexan 2+ Year Member

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Oh ok, i was wrong earlier than in saying that if there was friction or if it was connected to the spring, it would still decrease in velocity. Thought about it some more and i guess i am thinking about the rate of velocity changing (acceleration) when thinking of velocity. Furthermore i can visualize it now by thinking of a car accelerating fast in say 1st gear reaching 30 mph in a couple seconds..the Rate of velocity changing is higher, but the car is only going a maximum of 30 mph at its highest point. However once the car is in 6th gear and accelerating very slowly, the car is going say 70 mph it may have taken quite a few seconds or even minutes to go that fast, but its velocity is higher regardless because there was no negative acceleration. Wow, kinda sucks finding out such a glaring weakness soon before the test, but i am grateful and hopefully it saves me a point.. Thanks guys!

6. ### mehc012 Big Damn Hero 5+ Year Member

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Hey, the fact that you are able (and most importantly, willing) to read through, think it over, and actually notice oversights in logic is the key. I have seen way too many people 'study' by trying to prove themselves right, endlessly, and simply accepting and memorizing things when it is pointed out that the consensus differs from their opinion. Grappling with the info like you just did IS what gets high scores, and it is what will serve you well in all academic endeavors in the future. I was expecting a defensive and irritated retort, and your response was a pleasant surprise. I know I made my post blunt, but I find that it is best to avoid wiggle room to help avoid the behavior I described above.

We all have topics which our brains kind of skipped over and made assumptions on. The more you catch, the better off you are, no matter if it is 1month or 1d before the exam!

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### TheMightyTexan 2+ Year Member

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Thanks and agreed! With a little more than a week and a half left (sept 18) im trying to just get the rust off of old topics and re-review as much as i can.. i wish i had more time, but yes i definitely agree. Thanks again for the help, im sure ill have a few more questions on here haha.

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