Questions: A ball of mass 775 g hangs from a spring whose stiffness is 170 newtons per meter. A string is attached to the ball and you are pulling the string to the right, so that the ball hangs motionless, as shown in the figure. In this situation the spring is stretched, and its length is 15 cm. What would be the relaxed length of the spring, if it were detached from the ball and laid on a table?

Solution Steps

The problem involves a mass of 775 g hanging from a spring with a spring constant of 15 N/m. The mass is pulled down 5.5 cm and released. We need to determine the speed of the mass when it is 2.5 cm away from the equilibrium position.

Convert the mass and distances to standard units:

• Mass: 775 g = 0.775 kg
• Initial displacement: 5.5 cm = 0.055 m
• Final displacement: 2.5 cm = 0.025 m

The total mechanical energy in the system is conserved. The potential energy in the spring at the initial displacement is converted into kinetic energy and potential energy at the final displacement.

\[ PE_{initial} = \frac{1}{2} k x_{initial}^2 \] \[ PE_{initial} = \frac{1}{2} \times 15 \, \text{N/m} \times (0.055 \, \text{m})^2 \] \[ PE_{initial} = \frac{1}{2} \times 15 \times 0.003025 \] \[ PE_{initial} = 0.0226875 \, \text{J} \]

\[ PE_{final} = \frac{1}{2} k x_{final}^2 \] \[ PE_{final} = \frac{1}{2} \times 15 \, \text{N/m} \times (0.025 \, \text{m})^2 \] \[ PE_{final} = \frac{1}{2} \times 15 \times 0.000625 \] \[ PE_{final} = 0.0046875 \, \text{J} \]

The difference in potential energy is converted into kinetic energy: \[ KE = PE_{initial} - PE_{final} \] \[ KE = 0.0226875 \, \text{J} - 0.0046875 \, \text{J} \] \[ KE = 0.018 \, \text{J} \]

Use the kinetic energy to find the speed: \[ KE = \frac{1}{2} mv^2 \] \[ 0.018 \, \text{J} = \frac{1}{2} \times 0.775 \, \text{kg} \times v^2 \] \[ 0.018 = 0.3875 \times v^2 \] \[ v^2 = \frac{0.018}{0.3875} \] \[ v^2 = 0.0465 \] \[ v = \sqrt{0.0465} \] \[ v \approx 0.2156 \, \text{m/s} \]

Final Answer