Question 39

[student]

When the Joker was testing the spring, he applied a 700 N weight to the spring. By approximately how much was the spring compressed from its equilibrium position?


the answer is 6.6

however, the way the question was worded made me do the problem like this

originally, at equilibrium, we have to consider the platform

mg=kx
(300)=150x
x=2

then after 700 N is added
x = 6.67

so, after adding 700 N, it is displaced about 4.7 from equilibrium (6.67-2)

why is the answer 6.67?

[admin]

Well the answer 4.7 was one of the multiple choice options because there are some students who would make the same choice that you made.

I must admit, you probably will not be satisfied with this answer but here goes:

The real issue that you are questioning is the fact that the equilibrium position is not clearly defined: is it the true equilibrium position of the spring (you called it the original eq. posn)? Or, is it the new EQ posn which includes the platform?

It is ambiguous and I know that can be frustrating. But specifically for the MCAT, whenever there is ambiguity, take the simplest answer. In this situation, the question just wants the displacement from the true, original equilibrium position.