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Chapter 5: Problem 41

A spring with a spring constant of \(238.5 \mathrm{~N} / \mathrm{m}\) is compressed by \(0.231 \mathrm{~m}\). Then a steel ball bearing of mass \(0.0413 \mathrm{~kg}\) is put against the end of the spring, and the spring is released. What is the speed of the ball bearing right after it loses contact with the spring? (The ball bearing will come off the spring exactly as the spring returns to its equilibrium position. Assume that the mass of the spring can be neglected.)

Short Answer

Expert verified
Answer: The speed of the ball bearing right after it loses contact with the spring is approximately \(13.728\,\mathrm{m/s}\).

Step by step solution

01

Calculate the potential energy stored in the spring when compressed

We are given the spring constant (k) as \(238.5\,\mathrm{N/m}\), and the compression distance (x) as \(0.231\,\mathrm{m}\). We will use Hooke's law to find the potential energy (PE) stored in the compressed spring as follows: PE = \(0.5kx^2\) PE = \(0.5 \times 238.5\,\mathrm{N/m} \times (0.231\,\mathrm{m})^2\) = \(6.3800\,\mathrm{J}\).
02

Apply conservation of energy to find the kinetic energy of the ball bearing

As per the conservation of energy principle, the potential energy stored in the spring when compressed (PE) is converted to the kinetic energy (KE) gained by the ball bearing when it loses contact with the spring. So, KE = PE = \(6.3800\,\mathrm{J}\).
03

Calculate the speed of the ball bearing

Now, we'll use the mass (m) of the ball bearing to calculate its speed (v) right after it loses contact with the spring. The formula for kinetic energy is: KE = \(0.5mv^2\) Rearrange the formula to solve for the speed: \(v=\sqrt{\frac{2\,\text{KE}}{\text{m}}}\) Substitute the values for KE and m: \(v= \sqrt{\frac{2 \times 6.3800\,\mathrm{J}}{0.0413\,\mathrm{kg}}}\) \(v= 13.728\,\mathrm{m/s}\) So, the speed of the ball bearing right after it loses contact with the spring is approximately \(13.728\,\mathrm{m/s}\).

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