Spring Forces AP Physics 1 MCQ – Exam Style Questions etc.
Spring Forces AP Physics 1 MCQ
Unit: 2. Force and Translational Dynamics
Weightage : 10-15%
Exam Style Practice Questions, Spring Forces AP Physics 1 MCQ
Question
A block of mass M is suspended from two identical springs of negligible mass, spring constant kand unstretched length L. First, one spring is attached to the end of the other spring.
The block is then attached to the end of the second spring and slowly lowered to its equilibrium position. The two springs stretch a total distance of \(X_1\) , as shown in Figure 1 above. Next, the springs are hung side by side. The block is attached to the end of the springs and again slowly lowered to its equilibrium position. The springs each stretch a distance of \(X_2\) , as shown
in Figure 2 above. Which of the following equations correctly shows the relationship between \(X_1\) and \(X_2\) ?
(A)\(X_1=X_2\)
(B)\(X_1=\sqrt{2X_2}\)
(C)\(X_1=2X_2\)
(D)\(X_1=4X_2\)
(E)\(X_1=8X_2\)
Answer/Explanation
Ans:D
Question
In the figure above, the coefficient of static friction between the two blocks is 0.80. If the blocks oscillate with a frequency of 2.0 Hz, what is the maximum amplitude of the oscillations if the
small block is not to slip on the large block?
(A) 3.1 cm
(B) 5.0 cm
(C) 6.2 cm
(D) 7.5 cm
(E) 9.4 cm
Answer/Explanation
Ans: B
The horizontal position of the blocks can be given by the
equation
\( x = A cos(\omega t + \Phi _o)\)
Where A is the amplitude, \(\omega is the angular frequency, and \(\Phi _o\) is the initial phase. Differentiating this twice gives the acceleration:
\(a = \frac{d^2x}{dt^2}=-A\omega^2 cos (\omega t + \Phi _o) \Rightarrow a_{max} =A \omega^2\)
This means that the maximum force on block m is \(F_{max} = ma_{max} = mA \omega^2\). The static friction force must be able to provide this same amount of force; otherwise, the block will slip. Therefore,
\(F_{static,max} \geq m A \omega^2\)
\(\mu mg \geq m A \omega^2\)
\(A \leq \frac{\mu mg}{m \omega^2}=\frac{\mu g}{(2\pi f)^2}=\frac{(0.80)(10 m/s^2)}{4 \pi^2 (2.0 Hz)^2}\approx 0.05 m = 5.0 cm\)