3. A 2 kg mass is fixed to the end of a horizontal, frictionless spring $$ (\mathrm{K}=1000 \mathrm{~N} / \mathrm{m}) $$ which is ompressed 1.2 m . The mass is then released from rest. Determine the force acting on the spring when it $$ s 65 \mathrm{~cm} $$ from the spring equilibrium point. Determine the speed of the mass at this point as well.

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**Step 1. Determine the force acting on the spring**

The force provided by the spring follows Hooke’s law:
$$
F = -kx
$$
where  
$$ k = 1000 \, \mathrm{N/m} $$ (spring constant) and  
$$ x $$ is the displacement from the equilibrium position.

When the spring is compressed $$ 65 \, \mathrm{cm} $$ (or $$ 0.65 \, \mathrm{m} $$) from equilibrium, the magnitude of the force is
$$
|F| = kx = 1000 \times 0.65 = 650 \, \mathrm{N}.
$$
The negative sign in the formula indicates that the force acts in the direction opposite to the displacement.

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**Step 2. Determine the speed of the mass at this point**

The mass was initially compressed $$ 1.2 \, \mathrm{m} $$ and released from rest so that its initial energy is all stored as elastic potential energy.

The elastic potential energy stored in a spring is given by:
$$
U = \frac{1}{2} k x^2.
$$
- **Initial energy** when the spring is compressed $$ 1.2 \, \mathrm{m} $$:
  $$
  U_i = \frac{1}{2} \times 1000 \times (1.2)^2 = \frac{1}{2} \times 1000 \times 1.44 = 720 \, \mathrm{J}.
  $$

- **Energy at $$ x = 0.65 \, \mathrm{m} $$ from equilibrium**:
  $$
  U_f = \frac{1}{2} \times 1000 \times (0.65)^2 = \frac{1}{2} \times 1000 \times 0.4225 = 211.25 \, \mathrm{J}.
  $$

The decrease in potential energy has been converted into kinetic energy $$ (K) $$ of the mass:
$$
K = U_i - U_f = 720 \, \mathrm{J} - 211.25 \, \mathrm{J} = 508.75 \, \mathrm{J}.
$$

The kinetic energy of the mass is given by:
$$
K = \frac{1}{2} m v^2,
$$
where $$ m = 2 \, \mathrm{kg} $$. Solving for the speed $$ v $$:
$$
\frac{1}{2} m v^2 = 508.75 \quad \Longrightarrow \quad v^2 = \frac{2 \times 508.75}{2} = 508.75,
$$
$$
v = \sqrt{508.75} \approx 22.56 \, \mathrm{m/s}.
$$

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**Answer Summary**

- The magnitude of the force acting on the spring when it is $$ 65 \, \mathrm{cm} $$ from the equilibrium point is $$ 650 \, \mathrm{N} $$.
- The speed of the mass at that point is approximately $$ 22.56 \, \mathrm{m/s} $$.
When the spring is 65 cm from equilibrium, the force is 650 N. The mass's speed at that point is about 22.56 m/s.

A 2 kg mass is fixed to the end of a horizontal, frictionless spring
which is ompressed 1.2 m . The mass is then released from rest. Determine
the force acting on the spring when it from the spring equilibrium
point. Determine the speed of the mass at this point as well.

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A 2 kg mass is fixed to the end of a horizontal, frictionless spring which is ompressed 1.2 m . The mass is then released from rest. Determine the force acting on the spring when it from the spring equilibrium point. Determine the speed of the mass at this point as well.