Spring question (Hooke's law) E.K # 255

[Jack08]

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An object is placed on a spring. The spring is compressed downward and released. From the moment the spring is released until the object leaves the spring, the magnitude of the objects acceleration will:

A) Increase then decrease
B) decrease then increase

I chose the first one because when dealing with a spring on a friction-less surface acceleration does Increase and then Decrease. However, apparently when gravity is involved it is the opposite ( B. Decreases then Increases). I figured as the spring extended outward more, it would be relaxed more and more as x gets smaller and thus acceleration should Increase then Decrease. However, it's the opposite. Can someone explain? Thanks.

 

[NontradCA]

F=ma, a=F/m. F=kdeltax. As x decreases so does the force, and acceleration approaches 0. However, once it leaves the spring acceleration is due to gravity. I think you understand those but you have to realize the moment it is released, the acceleration is at it's max and only goes down from there.

 

[Jack08]

Because acceration is a vector.

the question asks about the magnitude. Thus direction is not needed when considering the answer....

 

[Jack08]

yes acceleration does approach 0 as you said. Which means the acceleration must first Increase and then slowly decrease as the distance of x gets smaller. However, the problem is saying that acceleration Decreases first. Then increases.

Just trying to talk this out. Truly appreciate the replies!

 

[Jack08]

Never-mind. I fully understand now. The explanation in the back was really confusing me the way they worded it. In my mind however, the object should be starting at an acceleration of 0 and then moving very quickly to its max acceleration in the first moments of release. (thus increasing first) and then as x decreases the acceleration should do the same (decreases) until it leaves the spring and only gravity is weighing on the spring in which case it should then increase again. But I guess they are ignoring the first increase. In any case. Thanks for your help!!

 

[gettheleadout]

Never-mind. I fully understand now. The explanation in the back was really confusing me the way they worded it. In my mind however, the object should be starting at an acceleration of 0 and then moving very quickly to its max acceleration in the first moments of release. (thus increasing first) and then as x decreases the acceleration should do the same (decreases) until it leaves the spring and only gravity is weighing on the spring in which case it should then increase again. But I guess they are ignoring the first increase. In any case. Thanks for your help!!

I totally understand where you're coming from here but the acceleration is never zero. I know that seems weird.

Think of an object thrown upward. At the apex its velocity is zero but the magnitude of its acceleration is g, right? That's where our spring-propelled object starts. The velocity plot from t=0 rises with a negative derivative, so the acceleration is decreasing at every point until the apex of the v(t) plot, which is NOT the apex of the object's flight but actually the moment it leaves the spring platform.

The moment you release the spring there is a non-zero force on the object, meaning it feels a non-zero acceleration. The acceleration decreases as it rises because the restoring force decreases, until the moment where F_r = F_g and the acceleration hits zero. (The a(t) graph is actually discontinuous at this point.) If the object had no momentum at this point, it would sit stationary on the spring, but since it has momentum it continues upward into the air. From that moment it is in free fall, and the acceleration is constant (equal to g).

The answer is a bit sketchy since the magnitude of the acceleration decreases to zero and then discontinuously increases to g. That is, the "increase" that follows the decrease occurs instantaneously. It's still right, just hard.