Question

A 5.0kg rock whose density is 4800 kg/m3 is suspended bya string such that half of the rock's volume is under water. What is the tension in the string?

Answer 1

Answer:

The tension in the string is 43.9 N.

Explanation:

Given that,

Mass of rock = 5.0 kg

Density of rock = 4800 kg/m³

We need to calculate the volume of rock

Using formula of density

$\rho=\dfrac{m}{V}$

$V=\dfrac{m}{\rho}$

Put the value into the formula

$V=\dfrac{5.0}{4800}$

$V=0.001041\ m^3$

We need to calculate the volume of water

$V_{w}=\dfrac{V}{2}$

Put the value of volume

$V_{w}=\dfrac{0.001041}{2}$

$V_{w}=0.0005205\ m^3$

We need to calculate the mass of water displaced

Using formula of mass

$m = 1000\times0.0005205$

$m=0.5205\ kg$

We need to calculate weight of water displaced

Using formula of weight of water

$W=0.5205\times9.8$

$W=5.1009\ N$

Weight of rock is

$W_{r}=5.0\times9.8$

$W_{r}=49\ N$

We need to calculate the tension in the string

Using formula of tension

$T=\text{weight of rock - weight of water displaced}$

Put the value into the formula

$T=49-5.1009$

$T=43.9\ N$

Hence, The tension in the string is 43.9 N.