Answer to Question #286347 in Physics for Kriz

Question

1. If a block of aluminum measures 4.0 cm x 5.0 cm x 2.0 cm is completely submerged in a tank of water, what volume of water does it displace?

2. What is the mass of the displaced water?

3. What is the weight of the displaced water?

4. How large of a buoyant force acts on the block?

5. The mass of the aluminum block is 108 g. Determine its density.

6. If you release the metal block, will it sink, or will it float to the surface? Explain your reasoning.

7. If the block was submerged in ethyl alcohol instead, would the tension in the string be bigger or smaller?

(With solution if it's okay with you, thank you so much)

Answer 1

1) The volume of water displaced by the block of aluminum equals to the volume of the block:

$$V_w=V_{block}=abc=4.0\ cm\times5.0\ cm\times2.0\ cm=40\ cm^3.$$

2) We can find the mass of the displaced water from the definition of the density:

$$\rho_w=\dfrac{m_w}{V_w},$$ $$m_w=\rho_wV_w=1.0\ \dfrac{g}{cm^3}\times40\ cm^3=40\ g=0.04\ kg.$$

3) We can find the weight of the displaced water as follows:

$$W_w=m_wg=0.04\ kg\times9.8\ \dfrac{m}{s^2}=0.4\ N.$$

4) Since the buoyant force that acts on the block is equal to the weight of the displaced water, we can write:

$$F_B=W_w=0.4\ N.$$

5) By the definition of the density, we have:

$$\rho_{block}=\dfrac{m_{block}}{V_{block}}=\dfrac{108\ g}{40\ cm^3}=2.7\ \dfrac{g}{cm^3}.$$

6) Let's first find the weight of the block:

$$W_{block}=m_{block}g=0.108\ kg\times9.8\ \dfrac{m}{s^2}=1.06\ N.$$

Since $W_{block}>F_B$, the block will sink.

7) Let's first find the tension in the string if the block submerged in water:

$$T+F_B-W_{block}=0,$$ $$T=W_{block}-F_B=1.06\ N-0.4\ N=0.66\ N.$$

In the case of ethyl alcohol we have:

$$T=W_{block}-\rho_wV_wg,$$ $$T=1.06\ N-785.1\ \dfrac{kg}{m^3}\times4\times10^{-5}\ m^3\times9.8\ \dfrac{m}{s^2}=0.75\ N.$$

Therefore, the tension in the string be bigger.