A \(230.0-\mathrm{g}\) object on a spring oscillates left to right on a frictionless surface with a frequency of \(2.00 \mathrm{Hz}\). Its position as a function of time is given by \(x=(8.00 \mathrm{cm})\) sin \(\omega t\) (a) Sketch a graph of the elastic potential energy as a function of time. (b) The object's velocity is given by \(v_{x}=\omega(8.00 \mathrm{cm}) \cos \omega t .\) Graph the system's kinetic energy as a function of time. (c) Graph the sum of the kinetic energy and the potential energy as a function of time. (d) Describe qualitatively how your answers would change if the surface weren't frictionless.

Short Answer

Expert verified
Answer: The presence of friction causes the amplitudes of the potential and kinetic energy graphs to diminish over time due to energy dissipation. The potential and kinetic energy graphs would still be out of phase, but their amplitudes would decrease as energy is converted into thermal energy by friction. The total energy graph would show a slow decrease in energy over time, as the total mechanical energy is no longer conserved due to the presence of friction.

Step by step solution

01

Calculate elastic potential energy as a function of time

We are given that the position of the mass is \(x = (8.00\,\mathrm{cm})\sin\omega t\). In order to find the elastic potential energy, we need to find the spring constant \(k\). We can use the formula for the frequency of a spring-mass system: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\). First, we can solve for the spring constant \(k\): \(f = 2.00 \mathrm{Hz}\), \(m = 230.0\,\mathrm{g} = 0.230\,\mathrm{kg}\) Solve for \(k\): \(k = (2\pi f)^2m = 58.9\,\mathrm{N/m}\) Now, we can calculate the potential energy using the formula \(U = \frac{1}{2}kx^2\): \(U(t) = \frac{1}{2}(58.9\,\mathrm{N/m})[(8.00\,\mathrm{cm})\sin(\omega t)]^2\)
02

Calculate kinetic energy as a function of time

We are given the velocity as \(v_{x}=\omega(8.00\,\mathrm{cm})\cos\omega t\). We can calculate the kinetic energy as \(K = \frac{1}{2}mv^2\): \(K(t) = \frac{1}{2}(0.230\,\mathrm{kg})[\omega(8.00\,\mathrm{cm})\cos\omega t]^2\)
03

Graph potential energy, kinetic energy, and total energy as functions of time

To graph the energies, first calculate the period \(T = \frac{1}{f} = 0.5\,s\) and the angular frequency \(\omega = 2\pi f = 12.6 \,\mathrm{rad/s}\): 1. Elastic potential energy: Graph the function \(U(t) = \frac{1}{2}(58.9\,\mathrm{N/m})[(8.00\,\mathrm{cm})\sin(12.6\,\mathrm{rad/s}\cdot t)]^2\) 2. Kinetic energy: Graph the function \(K(t) = \frac{1}{2}(0.230\,\mathrm{kg})[12.6\,\mathrm{rad/s}(8.00\,\mathrm{cm})\cos(12.6\,\mathrm{rad/s}\cdot t)]^2\) 3. Total energy: Calculate the sum of the kinetic and potential energy: \(E(t) = U(t) + K(t)\) Notice that the potential and kinetic energy graphs are out of phase; when one is maximum, the other is minimum. The total energy graph is constant since energy is conserved in this frictionless system.
04

Discuss the effects of friction on the system

If the surface were not frictionless, some of the mechanical energy (potential and kinetic energy) would be converted into thermal energy due to friction, causing a decrease in total mechanical energy over time. The graphs of potential and kinetic energy would still be out of phase, but the amplitudes would diminish over time (due to energy dissipation). As a result, the total energy graph would show a slow decrease in energy. However, the qualitative behavior of the system would still follow similar patterns, with potential and kinetic energy oscillating but gradually decreasing due to energy dissipation.

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