A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Here's a free body diagram of this emerging moon rock: there is a buoyant force upwards due to the weight of the water displaced and this is the apparent weight exerted by some spring scale that's keeping it suspended in the water and then there's gravity pulling it straight down. So we have apparent weight W prime plus buoyant force—both upwards— equals the gravity downwards. And we can solve this for the buoyant force and that's useful because we are going to use it to solve for the volume of the rock and then we'll take the known mass, divide by that volume and we'll get the density. So we have buoyant force equals F g minus W prime after we subtract W prime from both sides and substituting for buoyant force using Archimedes principle saying that it's equal to the weight of the water displaced. So it's the density of water times the volume of water displaced which is equal to the volume of the rock because it's submerged and times g and that equals the weight of the rock in a vacuum which is it's mass times g minus the apparent weight of the rock m prime times g. And the g's cancel and we can divide both sides by ρ H2O—density of water— that gives the volume of the rock. And the density of the rock is the mass that we are told divided by this volume so that's mass multiplied by the reciprocal of this volume— instead of dividing by a fraction, we'll multiply by its reciprocal it's a lot cleaner to look at— so we have mass multiplied by density of water divided by the difference in the vacuum mass minus the apparent mass once submerged in water and we have 9.28 kilograms times 1.00 times 10 to the 3 kilograms per cubic meter—density of water— divided by 9.28 minus 6.18 kilograms and that gives 2.99 times 10 to the 4 kilograms per cubic meter is the density of the moon rock.