To solve the problem step by step, we will address both parts of the question: (a) finding the acceleration of the block towards the surface when it is released, and (b) calculating the time for the block to reach the surface.
### Part (a): Finding the Acceleration of the Block
1. **Identify the Forces Acting on the Block**:
- The weight of the block (W) acting downwards.
- The buoyant force (F_b) acting upwards.
2. **Calculate the Weight of the Block**:
- The weight of the block can be expressed as:
\[
W = \text{Density of the block} \times \text{Volume of the block} \times g
\]
- Let the density of the block be \(\rho = 0.4 \times 10^3 \, \text{kg/m}^3\) and \(g \approx 9.8 \, \text{m/s}^2\).
3. **Calculate the Buoyant Force**:
- The buoyant force can be expressed as:
\[
F_b = \text{Density of water} \times \text{Volume of the block} \times g
\]
- The density of water is approximately \(\rho_L = 10^3 \, \text{kg/m}^3\).
4. **Set Up the Equation of Motion**:
- The net force (F_net) acting on the block when it is released is given by:
\[
F_{\text{net}} = F_b - W
\]
- Substituting the expressions for weight and buoyant force:
\[
F_{\text{net}} = (\rho_L \cdot V \cdot g) - (\rho \cdot V \cdot g)
\]
- Factor out \(V \cdot g\):
\[
F_{\text{net}} = V \cdot g \cdot (\rho_L - \rho)
\]
5. **Calculate the Acceleration**:
- Using Newton's second law, \(F = m \cdot a\), where \(m\) is the mass of the block:
\[
a = \frac{F_{\text{net}}}{m}
\]
- Since \(m = \rho \cdot V\), we can write:
\[
a = \frac{V \cdot g \cdot (\rho_L - \rho)}{\rho \cdot V} = g \cdot \frac{\rho_L - \rho}{\rho}
\]
- Substitute the values:
\[
a = g \cdot \frac{(10^3 - 0.4 \times 10^3)}{0.4 \times 10^3}
\]
\[
a = g \cdot \frac{600}{400} = g \cdot \frac{3}{2}
\]
- Therefore, the acceleration is:
\[
a = \frac{3}{2} g
\]
- Substituting \(g \approx 9.8 \, \text{m/s}^2\):
\[
a \approx \frac{3}{2} \times 9.8 \approx 14.7 \, \text{m/s}^2
\]
### Part (b): Finding the Time to Reach the Surface
1. **Use the Equation of Motion**:
- The equation of motion for an object starting from rest is:
\[
s = \frac{1}{2} a t^2
\]
- Here, \(s = 2.9 \, \text{m}\) (the depth) and \(a = 14.7 \, \text{m/s}^2\).
2. **Rearranging the Equation**:
- Rearranging for \(t\):
\[
t^2 = \frac{2s}{a}
\]
\[
t = \sqrt{\frac{2s}{a}}
\]
3. **Substituting the Values**:
- Substitute \(s = 2.9 \, \text{m}\) and \(a = 14.7 \, \text{m/s}^2\):
\[
t = \sqrt{\frac{2 \times 2.9}{14.7}} = \sqrt{\frac{5.8}{14.7}} \approx \sqrt{0.394} \approx 0.63 \, \text{s}
\]
### Final Answers:
- (a) The acceleration of the block towards the surface is approximately \(14.7 \, \text{m/s}^2\).
- (b) The time taken for the block to reach the surface is approximately \(0.63 \, \text{s}\).
