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A few people told me to check out http://www.mcatquestionaday.com/ while I'm studying for the MCAT. I've did the past few and they seem good (gives me a good refresher of something I might have forgotten). Then I ran into today's questions (9/2/08):
A spring which does not behave by Hooke's law has a force of kx2 where k is a constant and x is the displacement. The spring is hung on the ceiling and a spring of mass m is atched to the bottom of the spring. If the spring is released when the displacement of the spring is equal to zero, what is the maximum downward displacement?
(a) (mg/k)1/2
(b) (3mg/k)1/2
(c) mg/k
(d) 3mg/k
And the explanation:
To answer this question, we must look at work done by the spring and work done by gravity. We know that when the spring is released, gravity does work on the mass-spring system. Similarly, the spring does work as well; to be particular, the spring does equal and opposite work of gravity. We know that work is force multiplied by displacement. Let us say the maximum displacement is the variable x. Therefore, the work that gravity does on the system is mgx, because mg is the gravitational force and x is the displacement. Now we must calculate the work done by the spring. It is easy to make the mistake and say that the work done by the spring is kx3, by simply taking the force (kx2) and multiplying by displacement (x). This is NOT allowed because the spring does not follow Hooke's law. There is not a constant increase in force as the spring is stretched, rather the force increases quadradically. This means that you must use the integral and integrate over kx2 from the displacement of 0 to the displacement of x. This integral evaluates to kx3/3. Since we know that work done by the spring and work done by gravity are equivalent at the equilibrium point, we can set kx3/3 = mgx and solving for x yields that x = (3mg/k)1/2.
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Integrals...Seriously?? I thought the MCAT was still calculus free or are there tricks to answering questions using calculus (never really thought it would help before)? I really wouldn't even thought about integrals for this problem. Is this the type of q I can expect on the MCAT?
A spring which does not behave by Hooke's law has a force of kx2 where k is a constant and x is the displacement. The spring is hung on the ceiling and a spring of mass m is atched to the bottom of the spring. If the spring is released when the displacement of the spring is equal to zero, what is the maximum downward displacement?
(a) (mg/k)1/2
(b) (3mg/k)1/2
(c) mg/k
(d) 3mg/k
And the explanation:
To answer this question, we must look at work done by the spring and work done by gravity. We know that when the spring is released, gravity does work on the mass-spring system. Similarly, the spring does work as well; to be particular, the spring does equal and opposite work of gravity. We know that work is force multiplied by displacement. Let us say the maximum displacement is the variable x. Therefore, the work that gravity does on the system is mgx, because mg is the gravitational force and x is the displacement. Now we must calculate the work done by the spring. It is easy to make the mistake and say that the work done by the spring is kx3, by simply taking the force (kx2) and multiplying by displacement (x). This is NOT allowed because the spring does not follow Hooke's law. There is not a constant increase in force as the spring is stretched, rather the force increases quadradically. This means that you must use the integral and integrate over kx2 from the displacement of 0 to the displacement of x. This integral evaluates to kx3/3. Since we know that work done by the spring and work done by gravity are equivalent at the equilibrium point, we can set kx3/3 = mgx and solving for x yields that x = (3mg/k)1/2.
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Integrals...Seriously?? I thought the MCAT was still calculus free or are there tricks to answering questions using calculus (never really thought it would help before)? I really wouldn't even thought about integrals for this problem. Is this the type of q I can expect on the MCAT?