1. **State the problem:** We have a mass-spring system with mass $m=4$ kg and spring constant $k=100$ N/m. The mass is displaced by $0.2$ m from equilibrium and released from rest. We want to find the speed and total mechanical energy when the displacement is $x=0.1$ m. 2. **Relevant formulas:** - Total mechanical energy in a mass-spring system is constant and given by $$E = \frac{1}{2} k A^2$$ where $A$ is the amplitude of oscillation. - The speed $v$ at displacement $x$ is given by conservation of energy: $$E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2$$ 3. **Calculate total mechanical energy:** Given amplitude $A=0.2$ m, $$E = \frac{1}{2} \times 100 \times (0.2)^2 = 50 \times 0.04 = 2 \text{ J}$$ 4. **Find speed at $x=0.1$ m:** Using energy conservation, $$2 = \frac{1}{2} \times 100 \times (0.1)^2 + \frac{1}{2} \times 4 \times v^2$$ Simplify the potential energy term: $$\frac{1}{2} \times 100 \times 0.01 = 0.5$$ So, $$2 = 0.5 + 2 v^2$$ Subtract 0.5 from both sides: $$2 - 0.5 = 2 v^2$$ $$1.5 = 2 v^2$$ Divide both sides by 2: $$\cancel{2} v^2 = \frac{1.5}{\cancel{2}}$$ $$v^2 = 0.75$$ Take the square root: $$v = \sqrt{0.75} = 0.866 \text{ m/s}$$ 5. **Final answers:** - Speed at $x=0.1$ m is approximately $0.866$ m/s. - Total mechanical energy is $2$ J.