A 0.500-kg cube connected to a light spring for which the force constant is 20.0 N/m oscillates on a horizontal, frictionless track. (a) Calculate the total energy of the system and the maximum speed of the cube if the amplitude of the motion is 3.00 cm.
Question : A 0.500-kg cube connected to a light spring for which the fo...
- Mass of the cube: 0.500 kg
- Force constant of the light spring: 20.0 N/m
- Amplitude of the motion: 3.00 cm
Step 1:
Use Equation 13.22 to find the total mechanical energy (E) of the system. This equation states that the total energy is equal to the sum of the kinetic energy (K) and the potential energy (U). In this case, we are given the spring constant (k) and the maximum displacement (A), so we can plug in those values to calculate E.
Step 2:
Since the cube is at x=0, we know that the potential energy is zero and the total energy is equal to the kinetic energy. We can use this information, along with the value of E we found in step 1, to solve for the maximum velocity (v_max) of the cube.
Step 3:
To find the velocity of the cube when the displacement is 2.00 cm, we can use Equation 13.23. This equation relates the displacement (x) to the maximum displacement (A), the spring constant (k), and the mass (m) of the cube. We can plug in the given values and solve for the velocity.
Step 4:
Finally, to compute the kinetic and potential energies of the system when the displacement is 2.00 cm, we can use the velocity we found in step 3. The kinetic energy can be calculated using the equation K = 1/2 m v^2, where m is the mass of the cube and v is the velocity. The potential energy can be calculated using the equation U = 1/2 k x^2, where x is the displacement.
Note that the sum of the kinetic and potential energies should be equal to the total mechanical energy of the system, as stated by the principle of conservation of energy.
Final Answer
Using Equation 13.22, we obtain
E =K+U=\frac{1}{2} h \mathrm{~A}^{2}=\frac{1}{2}(20.0 \mathrm{~N} / \mathrm{m})\left(3.00 \times 10^{-2} \mathrm{~m}\right)^{2} =9.00 \times 10^{-3} \mathrm{~J}When the cube is at x=0, we know that U=0 and E=\frac{1}{2} m v_{\max }^{2} ; therefore
\frac{1}{2} m v_{\max }^{2} =9.00 \times 10^{-3} \mathrm{~J} v_{\max } =\sqrt{\frac{18.0 \times 10^{-3} \mathrm{~J}}{0.500 \mathrm{~kg}}}=0.190 \mathrm{~m} / \mathrm{s}
(b) What is the velocity of the cube when the displace
\text { ment is } 2.00 \mathrm{~cm} ?
We can apply Equation 13.23 directly:
y =\pm \sqrt{\frac{k}{m}\left(A^{2}-x^{2}\right)} =\pm \sqrt{\frac{20.0 \mathrm{~N} / \mathrm{m}}{0.500 \mathrm{~kg}}\left[(0.0300 \mathrm{~m})^{2}-(0.0200 \mathrm{~m})^{2}\right]} =\pm 0.141 \mathrm{~m} / \mathrm{s}The positive and negative signs indicate that the cube could be moving to either the right or the left at this instant.
(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 \mathrm{~cm}.
Using the result of (b), we find that
K=\frac{1}{2} m v^{2}=\frac{1}{2}(0.500 \mathrm{~kg})(0.141 \mathrm{~m} / \mathrm{s})^{2}=5.00 \times 10^{-3} \mathrm{~J} U=\frac{1}{2} k_{x}^{2}=\frac{1}{2}(20.0 \mathrm{~N} / \mathrm{m})(0.0200 \mathrm{~m})^{2}=4.00 \times 10^{-3} \mathrm{~J}Note that K+U=E