Q4.18E - University Physics with Modern Physics

Question

(a) An ordinary flea has a mass of  $210\mu g$ . How many newtons does it weigh? 

(b) The mass of a typical froghopper is 12.3 mg. How many newtons does it weigh? 

(c) A house cat typically weighs 45 N. How many pounds does it weigh, and what

is its mass in kilograms?

Step-by-Step Solution

Answer

(a) The weight of the $210\text{\hspace{0.17em}μg}$  flea is $2.06\times {10}^{-6}\text{\hspace{0.17em}N}$ .

(b) The weight of the 12.3 mg froghopper is $1.2\times {10}^{-4}\text{\hspace{0.17em}N}$.

(c) The mass of the cat weighing 45 N is 10.14 pounds or 4.6 kg.

Step 1: Given data

The mass of a flea is

  $\begin{array}{c}{m}_{fl}=210\text{\hspace{0.17em}μg}\\ =210\times \left(1\text{\hspace{0.17em}μg}\times \frac{{10}^{-9}\text{\hspace{0.17em}kg}}{1\text{\hspace{0.17em}μg}}\right)\\ =210\times {10}^{-9}\text{\hspace{0.17em}kg}\end{array}$

 

The mass of a froghopper is

 $\begin{array}{c}{m}_{fr}=12.3\text{\hspace{0.17em}mg}\\ =12.3\times \left(1\text{\hspace{0.17em}mg}\times \frac{{10}^{-6}\text{\hspace{0.17em}kg}}{1\text{\hspace{0.17em}μg}}\right)\\ =12.3\times {10}^{-6}\text{\hspace{0.17em}kg}\end{array}$

 

The weight of a house cat is

  $W_c=45\text{\hspace{0.17em}N}$

The acceleration due to gravity on Earth is

  $g=9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$

Step 2: Gravitational force

The downward gravitational force on a body of mass $m$  is

                 $F_g=mg$                                                                                                      …(1)

Step 3: Weight and mass of the animals

From equation (1), the weight of the flea is

  $\begin{array}{c}{W}_{fl}=m_{fl}g\\ =210\times {10}^{-9}\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\\ =2.06\times {10}^{-6}\times \left(1{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\times \frac{1\text{\hspace{0.17em}N}}{1{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}}\right)\\ =2.06\times {10}^{-6}\text{\hspace{0.17em}N}\end{array}$

The weight is  $2.06\times {10}^{-6}\text{\hspace{0.17em}N}$.

 

From equation (1), the weight of the froghopper is

  $\begin{array}{c}{W}_{rl}=m_{fr}g\\ =12.3\times {10}^{-6}\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\\ =1.2\times {10}^{-4}\times \left(1{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\times \frac{1\text{\hspace{0.17em}N}}{1{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}}\right)\\ =1.2\times {10}^{-4}\text{\hspace{0.17em}N}\end{array}$

The weight is $1.2\times {10}^{-4}\text{\hspace{0.17em}N}$$1.2\times {10}^{-4}\text{\hspace{0.17em}N}$.

From equation (1), the mass of the cat is

  $\begin{array}{c}{m}_c=\frac{W_c}{g}\\ =\frac{45\text{\hspace{0.17em}N}}{9.8{\text{\hspace{0.17em}m/s}}^2}\\ =4.6\cdot \left(\frac{1}{{\text{m/s}}^2}\right)\cdot \left(1\text{\hspace{0.17em}N}\times \frac{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^2}{1\text{\hspace{0.17em}N}}\right)\\ =4.6\text{\hspace{0.17em}kg}\end{array}$

 

The mass of the cat in pounds is

  $\begin{array}{c}{m}_c=4.6\text{\hspace{0.17em}kg}\\ =4.6\cdot \left(1\text{\hspace{0.17em}kg}\times \frac{\text{2.20462\hspace{0.17em}pounds}}{1\text{\hspace{0.17em}kg}}\right)\\ =10.14\text{\hspace{0.17em}pounds}\end{array}$

 

Thus, the mass if 10.14 pounds or 4.6 kg.