Question

a vertical container with base area measuring 14.0 cm by 17.0 cm is being filled with identical pieces of candy,each with a volume of 50.0 millimeter cube and mass of 0.0200 g assume that the volume of the empty spaces between the candies negligible. if the height of candies in the container increases at the rate of 0.250 cm/s at what rate kilograms per minute does the mass of the candies in the container increases?

Answer 1

Task. A vertical container with base area measuring $a=14.0$ cm by $b=17.0$ cm is being filled with identical pieces of candy, each with a volume of $V_{0}=50.0$ millimeter cube and mass of $m_{0}=0.0200g$. Assume that the volume of the empty spaces between the candies is negligible. If the height of candies in the container increases at the rate of $r=0.250$ cm/s at what rate kilograms per minute does the mass of the candies in the container increases?

Solution. Notice that the area of the base of the vertical container is equal to

$A=ab.$

Also the density of one candy is

$\rho=\frac{m_{0}}{V_{0}}.$

By assumption the height of the contained increases at the rate $r=0.250$ cm/s, so we can assume that the height $h(t)$ at time $t$ is given by the following rule:

$h(t)=rt.$

The volume of the empty spaces between the candies is negligible, so we can assume that at time $t$ the volume of candies in the container is equal to

$V(t)=Ah(t)=abrt.$

Therefore the mass of the candies in the container at time $t$ is

$m(t)=\rho V(t)=\rho abrt=\frac{m_{0}abr}{V_{0}}t.$

Hence the rate of the mass of candies in the container is equal to

$m^{\prime}(t)=\left(\frac{m_{0}abr}{V_{0}}t\right)^{\prime}=\frac{m_{0}abr}{V_{0}}.$

Substituting values we get

$m^{\prime}$ $=\frac{m_{0}abr}{V_{0}}=\frac{0.02\ g\cdot 14\ cm\cdot 17\ cm\cdot 0.25\ cm/s}{50\ mm^{3}}=\frac{0.02\ g\cdot 14\ cm\cdot 17\ cm\cdot 0.25\ cm/s}{50\ mm^{3}}$

$=\frac{1.19\ g\cdot cm^{3}/s}{50\cdot 0.1^{3}\ cm^{3}}=23.8g/s=\frac{23.8\cdot 0.001\ kg}{1/3600\ hours}=85.68\ kg/hour.$

Answer. $m^{\prime}=\frac{m_{0}abr}{V_{0}}=85.68\ kg/hour.$