Unit 2: Force and Translational Dynamics
Weightage : 20-15%
Systems A and B contain identical ideal springs and identical blocks that can slide along a surface of negligible friction. In system A, the surface is horizontal. In system B, the surface makes an angle \(\theta\) with the horizontal.
Initially, both blocks are at rest and in equilibrium. Each block is then pulled the same distance \(d\) in the direction shown in the figures and released from rest at \(t=0\).
A student wants to use an apparatus similar to system B to measure the acceleration due to gravity \(g\). If the mass of the block, the spring constant, and the angle of the incline are known, what additional data must be measured to determine an experimental value for \(g\)?
I. The stretch of the spring at the equilibrium position
II. The speed of the block as it passes the equilibrium position
III. The time interval between two consecutive passes through the equilibrium position
Correct Answer: \( \boxed{\mathrm{A}} \)
At the equilibrium position, the net force on the block is zero. Along the incline,
\(F_{\mathrm{spring}}=F_{\mathrm{gravity}}\)
\(k\Delta x=mg\sin\theta\)
Solving for the acceleration due to gravity gives
\(g=\dfrac{k\Delta x}{m\sin\theta}\)
Since \(m\), \(k\), and \(\theta\) are already known, the only additional quantity that must be measured is the equilibrium stretch \(\Delta x\) of the spring.
The speed of the block at equilibrium and the time between consecutive passes through equilibrium depend on the oscillatory motion, but neither is needed to determine \(g\).
Therefore, only statement I is required.
Therefore, the correct answer is (A).
The figure shows blocks with different masses hanging from a spring or combination of two springs. All the springs are ideal and have spring constants of \(k\).
Which of the following correctly ranks the systems in order of most stored elastic potential energy to least stored elastic potential energy when the spring systems are stretched a distance \(d\) from their unstretched length?
Correct Answer: \( \boxed{\mathrm{E}} \)
The elastic potential energy stored in a spring system is
\(U=\frac{1}{2}k_{\mathrm{eq}}d^2\)
where \(k_{\mathrm{eq}}\) is the equivalent spring constant and \(d\) is the total extension from the unstretched length.
Determine the equivalent spring constant for each system:
System 1: One spring
\(k_{\mathrm{eq}}=k\)
System 2: Two identical springs in series
\(k_{\mathrm{eq}}=\dfrac{k}{2}\)
System 3: Two identical springs in parallel
\(k_{\mathrm{eq}}=2k\)
Therefore,
\(U_3=\frac{1}{2}(2k)d^2=kd^2\)
\(U_1=\frac{1}{2}kd^2\)
\(U_2=\frac{1}{2}\left(\frac{k}{2}\right)d^2=\frac{1}{4}kd^2\)
Thus,
\(U_3>U_1>U_2\)
Hence, the correct ranking is
\(\boxed{3>1>2}\)
Therefore, the correct answer is (E).
A particle is moving along the \(x\)-axis under the influence of a spring force \(F_x\). The potential energy \(U\) of the particle as a function of position \(x\) is given by
\(U(x)=\beta x^4\),
where
\(\beta=\dfrac{1}{2}\,\dfrac{\mathrm{J}}{\mathrm{m}^4}\).
The magnitude of \(F_x\) when the particle is at the position \(x=0.10\,\mathrm{m}\) is most nearly
Correct Answer: \( \boxed{\mathrm{C}} \)
For a conservative spring force,
\(F_x=-\dfrac{dU}{dx}\)
Differentiate the potential energy function:
\(F_x=-\dfrac{d}{dx}\left(\beta x^4\right)=-4\beta x^3\)
Substituting \(\beta=\dfrac{1}{2}\,\dfrac{\mathrm{J}}{\mathrm{m}^4}\),
\(F_x=-4\left(\dfrac{1}{2}\right)x^3=-2x^3\)
At \(x=0.10\,\mathrm{m}\),
\(F_x=-2(0.10)^3=-2(0.001)=-2\times10^{-3}\,\mathrm{N}\)
Therefore, the magnitude of the force is
\(|F_x|=2\times10^{-3}\,\mathrm{N}\)
Therefore, the correct answer is (C).
