A rectangular box of negligible mass measures 5m by 1m by .5m. How many kgs of mass can be loaded onto the box before it sinks in the water?
V=5*1*.5 = 2.5m^3
density of water =1g/cm^3 =1000 kg/m^3
I also know that when the density of the mass is greater than the density of water then the mass will sink. How would I solve this problem? Thanks in advance for your help.
To solve this problem, you need to compare the density of the box with the density of water. The density is calculated by dividing the mass by the volume.
1. Start by calculating the volume of the box:
- The length, width, and height of the box are given: 5m, 1m, and 0.5m, respectively.
- The formula for volume is V = length * width * height.
- So, V = 5m * 1m * 0.5m = 2.5m^3.
2. The density of water is given as 1000 kg/m^3. This means that 1 cubic meter of water weighs 1000 kilograms.
3. Next, calculate the maximum mass the box can hold without sinking into the water:
- Since the box's mass is negligible, we can assume all the mass is due to the load.
- The load's mass is equal to the density of water multiplied by the volume displaced by the box.
- So, mass = density * volume = 1000 kg/m^3 * 2.5m^3 = 2500 kg.
Therefore, the box can hold a maximum of 2500 kilograms of mass before it sinks in the water.