problem: a block of mass $m = 0.88\\text{ kg}$ is connected to a spring of force constant $k = 845\\text{…

problem: a block of mass $m = 0.88\\text{ kg}$ is connected to a spring of force constant $k = 845\\text{ n/m}$ on a smooth, horizontal surface. plot the potential energy of the spring from $x = -5.00\\text{ cm}$ to $x = 5.00\\text{ cm}$.

problem: a block of mass $m = 0.88\\text{ kg}$ is connected to a spring of force constant $k = 845\\text{ n/m}$ on a smooth, horizontal surface. plot the potential energy of the spring from $x = -5.00\\text{ cm}$ to $x = 5.00\\text{ cm}$.

Answer

Explanation:

Step1: Identify the potential energy formula

The elastic potential energy $U$ of a spring is given by: $$U(x) = \frac{1}{2} k x^2$$

Step2: Convert units to SI

Convert the displacement range from centimeters to meters: $$x_{min} = -5.00\text{ cm} = -0.05\text{ m}$$ $$x_{max} = 5.00\text{ cm} = 0.05\text{ m}$$

Step3: Calculate energy at boundaries

Substitute $k = 845\text{ N/m}$ and $x = \pm 0.05\text{ m}$ into the formula: $$U(\pm 0.05) = \frac{1}{2} (845) (0.05)^2$$ $$U(\pm 0.05) = 0.5 \times 845 \times 0.0025 = 1.05625\text{ J}$$

Step4: Determine key plotting points

Calculate $U$ for several points to define the parabolic shape:

  • At $x = 0$, $U = 0\text{ J}$
  • At $x = \pm 0.025\text{ m}$, $U = \frac{1}{2}(845)(0.025)^2 \approx 0.264\text{ J}$

Answer:

The plot is a parabola opening upward with the vertex at the origin $(0,0)$. The horizontal axis represents displacement $x$ from $-5.00\text{ cm}$ to $5.00\text{ cm}$, and the vertical axis represents potential energy $U$ in Joules (J). The function to plot is $U(x) = 422.5 x^2$ (where $x$ is in meters). Key points for the graph are $(-5, 1.06)$, $(0, 0)$, and $(5, 1.06)$ when $x$ is expressed in cm and $U$ in J.