Certainly! Let's work through this step-by-step:
1. Given Values:
- Volume of the cube, [tex]\( V = 0.2 \text{ m}^3 \)[/tex]
- Density of the cube, [tex]\( \rho_{\text{cube}} = 600 \text{ kg/m}^3 \)[/tex]
- Density of the liquid, [tex]\( \rho_{\text{liquid}} = 800 \text{ kg/m}^3 \)[/tex]
2. Gravitational Acceleration (g):
- [tex]\( g = 9.81 \text{ m/s}^2 \)[/tex]
3. Calculate the Weight of the Cube:
The weight of the cube, [tex]\( W_{\text{cube}} \)[/tex], can be calculated using the formula:
[tex]\[
W_{\text{cube}} = V \times \rho_{\text{cube}} \times g
\][/tex]
Substituting the given values:
[tex]\[
W_{\text{cube}} = 0.2 \times 600 \times 9.81 = 1177.2 \text{ N}
\][/tex]
4. Calculate the Buoyant Force:
The buoyant force, [tex]\( F_{\text{buoyant}} \)[/tex], is equal to the weight of the liquid displaced by the cube. This can be calculated using the formula:
[tex]\[
F_{\text{buoyant}} = V \times \rho_{\text{liquid}} \times g
\][/tex]
Substituting the given values:
[tex]\[
F_{\text{buoyant}} = 0.2 \times 800 \times 9.81 = 1569.6 \text{ N}
\][/tex]
5. Calculate the Force that Must Be Applied:
To keep the top surface of the cube at the same level as the liquid surface, the net force acting on the cube should be zero. This means the weight of the cube must be balanced by the buoyant force plus any applied force. The applied force [tex]\( F_{\text{applied}} \)[/tex] can be calculated as:
[tex]\[
F_{\text{applied}} = F_{\text{buoyant}} - W_{\text{cube}}
\][/tex]
Substituting the values we have calculated:
[tex]\[
F_{\text{applied}} = 1569.6 - 1177.2 = 392.4 \text{ N}
\][/tex]
Therefore, the force that must be applied to the cube to keep its top surface at the same level as the liquid surface is [tex]\( 392.4 \text{ N} \)[/tex].
