Luke is trying to catch a pesky animal that keeps eating vegetables from his garden. He is building a trap and needs to use a spring to close the door to his trap. He has a spring in his garage and he wants to determine the spring constant of the spring. To do this, he hangs the spring from the ceiling and measures that it is \(20.0 \mathrm{cm}\) long. Then he hangs a \(1.10-\mathrm{kg}\) brick on the end of the spring and it stretches to $31.0 \mathrm{cm} .$ (a) What is the spring constant of the spring? (b) Luke now pulls the brick \(5.00 \mathrm{cm}\) from the equilibrium position to watch it oscillate. What is the maximum speed of the brick? (c) When the displacement is \(2.50 \mathrm{cm}\) from the equilibrium position, what is the speed of the brick? (d) How long will it take for the brick to oscillate five times?

Step by step solution

01

(a) Determine the spring constant

To find the spring constant (k) of the spring, we will use Hooke's Law, which states that the force exerted by a spring (F) is proportional to its elongation (x): \(F = -kx\). In this case, the force acting on the spring is the gravitational force, which can be calculated as \(F = mg\). By equating both expressions of the force, we have \(mg = kx\), and thus we can solve for k: \(k = \frac{mg}{x}\). Using the given data, we can calculate the spring constant, where \(m = 1.10\,\mathrm{kg}\), \(g = 9.81\,\mathrm{m/s^2}\), and \(x = (31.0 - 20.0)\times 10^{-2}\,\mathrm{m}\): \(k = \frac{(1.10)(9.81)}{0.11} = 98.39\,\mathrm{N/m}\).

02

(b) Calculate the maximum speed

To determine the maximum speed of the brick, we will apply the conservation of mechanical energy principle. At its equilibrium position, the spring energy is completely converted into kinetic energy. Thus, the equation becomes: \(U_{sp} = \frac{1}{2} m v_{max}^2\) The potential energy stored in the spring at its maximum elongation (5 cm) is given by: \(U_{sp} = \frac{1}{2}kx_{max}^2\). Plugging in the values, we have \(U_{sp} = \frac{1}{2}(98.39)(0.05)^2 = 0.123\,\mathrm{J}\). Now, we can solve for \(v_{max}\): \(\frac{1}{2} m v_{max}^2 = 0.123\,\mathrm{J}\), giving us \(v_{max} = \sqrt{\frac{2(0.123)}{1.1}} = 0.475\,\mathrm{m/s}\).

03

(c) Determine the speed at a given displacement

At a given displacement of \(2.50\,\mathrm{cm}\) from the equilibrium position, we need to find the brick's speed. We'll again use the conservation of mechanical energy principle: \(U_{sp} + K = U_{sp_0} + K_0\). At this given displacement, we have both potential and kinetic energy, whereas at the maximum elongation, all the energy is stored in the spring. The equation becomes: \(\frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kx_{max}^2\). Plugging in the known values, we have: \(\frac{1}{2}(98.39)(0.025)^2 + \frac{1}{2}(1.1)v^2 = \frac{1}{2}(98.39)(0.05)^2\). Solving for \(v\), we get \(v = \sqrt{\frac{(98.39)(0.05^2 - 0.025^2)}{1.1}} = 0.398\,\mathrm{m/s}\).

04

(d) Calculate the time to oscillate five times

To determine the time required for the brick to oscillate five times, we need to find the oscillation period (T) and multiply it by 5. The period of oscillation for a spring-mass system is given by \(T = 2\pi\sqrt{\frac{m}{k}}\). Plugging in the values, we have \(T = 2\pi\sqrt{\frac{1.1}{98.39}} = 0.667\,\mathrm{s}\). Now, we can calculate the time for the brick to oscillate five times: \(t = 5T = 5(0.667) = 3.33\,\mathrm{s}\).

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