Unit 7: Oscillations
Weightage : 20-15%
A block of mass \(m\) is on a rough horizontal surface and is attached to a spring with spring constant \(k\). The coefficient of kinetic friction between the surface and the block is \(\mu\). When the block is at position \(x=0\), the spring is at its unstretched length. The block is pulled to position \(x=+x_0\), as shown above, and released from rest. The block then travels to the left and passes through \(x=0\) before coming momentarily to rest at position \(x=-\dfrac{x_0}{2}\).
Which of the following is a correct expression for the coefficient of kinetic friction \(\mu\)?
Correct Answer: \( \boxed{\mathrm{A}} \)
Initially, the spring stores elastic potential energy
\(U_i=\dfrac{1}{2}kx_0^2\)
When the block comes to rest at \(x=-\dfrac{x_0}{2}\), the remaining spring potential energy is
\(U_f=\dfrac{1}{2}k\left(\dfrac{x_0}{2}\right)^2=\dfrac{1}{8}kx_0^2\)
The energy dissipated by friction equals the decrease in mechanical energy:
\(W_f=U_i-U_f=\dfrac{1}{2}kx_0^2-\dfrac{1}{8}kx_0^2=\dfrac{3}{8}kx_0^2\)
The block travels a total distance
\(d=x_0+\dfrac{x_0}{2}=\dfrac{3x_0}{2}\)
The work done by friction is
\(W_f=\mu mgd=\mu mg\left(\dfrac{3x_0}{2}\right)\)
Equating the two expressions for the work done by friction,
\(\mu mg\left(\dfrac{3x_0}{2}\right)=\dfrac{3}{8}kx_0^2\)
Solving for \(\mu\),
\(\boxed{\mu=\dfrac{kx_0}{4mg}}\)
Therefore, the correct answer is (A).
A block of mass \(m\) is on a rough horizontal surface and is attached to a spring with spring constant \(k\). The coefficient of kinetic friction between the surface and the block is \(\mu\). When the block is at position \(x=0\), the spring is at its unstretched length. The block is pulled to position \(x=+x_0\), as shown above, and released from rest. The block then travels to the left and passes through \(x=0\) before coming momentarily to rest at position \(x=-\dfrac{x_0}{2}\).
Which of the following is a correct expression for the kinetic energy of the block as it first travels through \(x=0\)?
Correct Answer: \( \boxed{\mathrm{C}} \)
Initially, the block is released from rest at \(x=x_0\), so all of the mechanical energy is stored as elastic potential energy:
\(U_{s,i}=\dfrac{1}{2}kx_0^2\)
As the block moves from \(x=x_0\) to \(x=0\), kinetic friction acts over a distance \(x_0\). The work done by friction is
\(W_f=\mu mgx_0\)
Applying the work-energy principle,
\(K=U_{s,i}-W_f\)
Therefore,
\(K=\dfrac{1}{2}kx_0^2-\mu mgx_0\)
This is the kinetic energy of the block the first time it passes through the equilibrium position.
Therefore, the correct answer is (C).
A \(2.00\,\mathrm{kg}\) block is attached to a horizontal ideal spring with spring constant \(k=100\,\mathrm{N/m}\). The block-spring system is on a horizontal surface with negligible friction. A graph of the potential energy \(U\) as a function of time \(t\) for this system is shown.
The maximum displacement \(x_{\max}\) of the block from its equilibrium position and the maximum speed \(v_{\max}\) of the block during the motion represented by the graph are most nearly
Correct Answer: \( \boxed{\mathrm{C}} \)
From the graph, the maximum spring potential energy is approximately
\(U_{\max}=2.0\,\mathrm{J}\)
The spring potential energy is
\(U=\frac{1}{2}kx^2\)
Therefore,
\(x_{\max}=\sqrt{\frac{2U_{\max}}{k}} =\sqrt{\frac{2(2.0)}{100}} =0.20\,\mathrm{m}\)
At the equilibrium position, all of the spring potential energy has been converted into kinetic energy:
\(K_{\max}=U_{\max}=2.0\,\mathrm{J}\)
Using
\(K=\frac{1}{2}mv^2\)
\(2.0=\frac{1}{2}(2.00)v_{\max}^2\)
\(v_{\max}=\sqrt{2.0}\approx1.41\,\mathrm{m/s}\)
Thus,
\(x_{\max}=0.20\,\mathrm{m}\) and \(v_{\max}=1.4\,\mathrm{m/s}\).
Therefore, the correct answer is (C).
