Given Information:

• Spring constant: $k = 200$ N/m
• Compression: $x = 0.15$ m

(a) Find spring force

Use Hooke's Law:

$F = kx$

(Using magnitude; direction is toward equilibrium)

$F = (200)(0.15) = 30 \text{ N}$

The force points outward (opposite to compression).

(b) Find elastic potential energy

$PE_{elastic} = \frac{1}{2}kx^2$

$PE_{elastic} = \frac{1}{2}(200)(0.15)^2$

$PE_{elastic} = 100(0.0225)$

$PE_{elastic} = 2.25 \text{ J}$

Answers:

• (a) Spring force: 30 N (directed outward, opposing compression)
• (b) Elastic PE stored: 2.25 J

Given Information:

• Mass: $m = 0.5$ kg
• Spring constant: $k = 100$ N/m
• Initial compression: $x_i = 0.2$ m
• Frictionless surface

Find: Maximum speed $v_{max}$

Analysis:

Maximum speed occurs at equilibrium position ($x = 0$) where all elastic PE converts to KE.

Step 1: Calculate initial energy (at maximum compression)

At $x = 0.2$ m:

• $KE_i = 0$ (released from rest)
• $PE_i = \frac{1}{2}kx_i^2 = \frac{1}{2}(100)(0.2)^2 = 50(0.04) = 2$ J

Total initial energy: $E_i = 2$ J

Step 2: Calculate final energy (at equilibrium)

At $x = 0$:

• $KE_f = \frac{1}{2}mv_{max}^2$ (unknown)
• $PE_f = \frac{1}{2}k(0)^2 = 0$ J

Total final energy: $E_f = \frac{1}{2}mv_{max}^2$

Step 3: Apply conservation of energy

$E_i = E_f$

$2 = \frac{1}{2}(0.5)v_{max}^2$

$2 = 0.25v_{max}^2$

$v_{max}^2 = 8$

$v_{max} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \text{ m/s}$

Alternative formula:

For a spring-mass system released from rest at maximum displacement:

$v_{max} = x_{max}\sqrt{\frac{k}{m}} = 0.2\sqrt{\frac{100}{0.5}} = 0.2\sqrt{200} = 0.2(10\sqrt{2}) = 2\sqrt{2}$ m/s ✓

Answer: The maximum speed is $2\sqrt{2}$ m/s or approximately 2.83 m/s.

Given Information:

• Mass: $m = 2$ kg
• Spring constant: $k = 500$ N/m
• Initial displacement: $x_i = 0.1$ m (down from equilibrium)
• Vertical spring

(a) Find total mechanical energy

Step 1: Choose reference for gravitational PE

Let equilibrium position be $h = 0$ for gravitational PE.

Step 2: Calculate energy at release point

At release (0.1 m below equilibrium):

• $KE_i = 0$ (released from rest)
• $PE_{elastic,i} = \frac{1}{2}kx_i^2 = \frac{1}{2}(500)(0.1)^2 = 250(0.01) = 2.5$ J
• $PE_{grav,i} = mgh_i = (2)(9.8)(-0.1) = -1.96$ J (negative because below equilibrium)

Total energy: $E = 0 + 2.5 + (-1.96) = 0.54 \text{ J}$

(b) Find maximum height above release point

Step 3: At maximum height

At highest point, block momentarily stops ($v = 0$) and spring returns through equilibrium and compresses.

Let's say spring compresses by distance $d$ above equilibrium.

At this point:

• $KE = 0$
• $PE_{elastic} = \frac{1}{2}kd^2$
• $PE_{grav} = mg(d)$ (height $d$ above equilibrium, which is $d + 0.1$ above release)

Step 4: Apply conservation of energy

$E_{initial} = E_{final}$

$0.54 = 0 + \frac{1}{2}kd^2 + mgd$

$0.54 = 250d^2 + 19.6d$

$250d^2 + 19.6d - 0.54 = 0$

Step 5: Solve quadratic

Using quadratic formula: $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$d = \frac{-19.6 \pm \sqrt{(19.6)^2 - 4(250)(-0.54)}}{2(250)}$

$d = \frac{-19.6 \pm \sqrt{384.16 + 540}}{500}$

$d = \frac{-19.6 \pm \sqrt{924.16}}{500}$

$d = \frac{-19.6 \pm 30.4}{500}$

Taking positive solution: $d = \frac{10.8}{500} = 0.0216 \text{ m} = 2.16 \text{ cm}$

Height above release point: $h = d + 0.1 = 0.0216 + 0.1 = 0.1216$ m

Simplified approach (if we ignore spring compression at top):

If spring just returns to equilibrium ($d = 0$):

$0.54 = 0 + 0 + mg(h)$ $h = \frac{0.54}{19.6} \approx 0.0276$ m

But spring actually compresses slightly, giving total height ≈ 0.122 m.

Answers:

• (a) Total mechanical energy: 0.54 J
• (b) Height above release point: approximately 0.122 m or 12.2 cm

Note: This problem is complex because both gravitational and elastic PE change. The spring compresses slightly above equilibrium before the block stops.