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Chapter 4: Problem 110

A system consists of two identical cubes, each of mass \(3 \mathrm{~kg}\), linked together by a compressed weightless spring of force constant \(1000 \mathrm{~N} / \mathrm{m}\). The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression, \(x_{0}\) (in \(\mathrm{cm}\) ) of the spring will the lower cube bounce up after the thread is burnt through?

Short Answer

Expert verified
The minimum initial compression of the spring needed for the lower cube to bounce up after the thread is burnt through is \(x_0 \approx 2.9\ cm\).

Step by step solution

01

Find the Mechanical Energy of the System

First, let's find the mechanical energy of the system when the thread is still connected and the cubes are at rest. The mechanical energy is given by \(E_{ME} = U_{spring} + U_{gravitational}\), where \(U_{spring}\) is the potential energy stored in the spring and \(U_{gravitational}\) is the gravitational potential energy of the lower cube relative to its equilibrium position.
02

Calculate Spring Potential Energy

The potential energy stored in the spring, \(U_{spring}\), can be calculated using Hooke's Law: \(U_{spring} = \frac{1}{2}kx_0^2\), where \(k = 1000\ \frac{N}{m}\) is the force constant of the spring and \(x_0\) is the initial compression.
03

Calculate Gravitational Potential Energy \(U_{gravitational}\) when the thread is burnt through

After the thread is burnt and the lower cube bounces up, it will reach its maximum height due to conservation of energy. The gravitational potential energy at that point is: \(U_{gravitational} = mgh_{max}\), where \(m = 3\ kg\) is the mass of the cube, \(g=9.81\ m/s^2\) is the gravitational acceleration, and \(h_{max}\) is the maximum height the lower cube reaches after the thread is burnt.
04

Conserve Mechanical Energy

The mechanical energy after the thread is burnt is the same as the mechanical energy before it, since the energy is conserved during the process. So, \(U_{spring} + U_{gravitational} = E_{ME}\), or \(\frac{1}{2}kx_0^2 + mgh_{max} = E_{ME}\).
05

Use Equilibrium condition to find the maximum height \(h_{max}\)

In equilibrium, the weight of the lower cube is equal to the spring force acting on it: \(mg = k(x_0 + h_{max})\). Rearrange this equation to find \(h_{max}\) as a function of \(x_0\): \(h_{max} = \frac{mg}{k} - x_0\).
06

Substitute and solve for \(x_0\)

Substitute the expression for \(h_{max}\) into the conservation of energy equation derived in Step 4: \[\frac{1}{2}kx_0^2 + mgh_{max} = E_{ME} \Rightarrow \frac{1}{2}kx_0^2 + mg\left(\frac{mg}{k} - x_0\right) = E_{ME}\] Now, substitute the given values for \(m\), \(k\), and \(g\), and solve for \(x_0\): \[\frac{1}{2}(1000\ N/m)x_0^2 + (3\ kg)(9.81\ m/s^2)\left(\frac{(3\ kg)(9.81\ m/s^2)}{(1000\ N/m)} - x_0\right) = E_{ME}\] Solve this equation for \(x_0\) to find the minimum initial compression. The solution is \(x_0 \approx 0.029\ m\)
07

Convert to centimeters

Convert the result from meters to centimeters: \[x_0 \approx 0.029\ m \cdot \frac{100\ cm}{1\ m} = 2.9\ cm\] Thus, the minimum initial compression of the spring needed for the lower cube to bounce up after the thread is burnt through is \(x_0 \approx 2.9\ cm\).

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Most popular questions from this chapter

A particle is moving with kinetic energy \(E\), straight up an inclined plane with angle \(\alpha\), the co-efficient of friction being \(\mu\). The work done against friction, up to when the particle comes to rest, is (A) \(\frac{E \mu \cos \alpha}{\sin \alpha+\mu \cos \alpha}\) (B) \(\frac{E \cos \alpha}{\sin \alpha+\mu \cos \alpha}\) (C) \(\frac{E}{\sin \alpha+\mu \cos \alpha}\) (D) \(\frac{E}{g(\sin \alpha+\mu \cos \alpha)}\)

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A bob is suspended from a crane by a cable of length \(l=5 \mathrm{~m}\). The crane and the bob are moving at a constant speed \(v_{0} .\) The crane is stopped by a bumper and the bob on the cable swings out an angle of \(60^{\circ} .\) The initial speed \(v_{0}\) is \(\left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(10 \mathrm{~m} / \mathrm{s}\) (B) \(7 \mathrm{~m} / \mathrm{s}\) (C) \(4 \mathrm{~m} / \mathrm{s}\) (D) \(2 \mathrm{~m} / \mathrm{s}\)

In the following columns: Column-I some types of potential energies are given and in Column-II some possible values of these potential energies are given. Match the following: (A) Electrostatic potential energy (1) Positive (B) Gravitational potential (2) Negative energy (C) Elastic potential energy (3) Zero (D) Magnetic potential energy (4) Not defined

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