Chapter 3, Problem 1TC

Compare from greatest to least, the accelerations of these objects:

(a) Net force $\text{2}0\text{N}$ acts on a mass of $10\text{kg}$ .

(b) Net force $\text{3}0\text{N}$ acts on a mass of $18\text{kg}$ .

(c) Net force $\text{4}0\text{N}$ acts on a mass of $30\text{kg}$ .

(d) Net force $\text{5}0\text{N}$ acts on a mass of $32\text{kg}$ .

Greatest ___ ___ ___ ___ Least

To determine

The ranking from greatest to least of acceleration of the objects.

Answer to Problem 1TC

The ranking from greatest to least of acceleration of the objects is ${\text{a}}_a>{\text{a}}_b{\text{>a}}_d>{\text{a}}_c.$

Explanation of Solution

Given info: Net force in case (a) is $\text{2}0\text{N}$ ,mass of the object in case (a) is $10\text{kg}$ , net force in case (b) is $\text{3}0\text{N}$ ,mass of the object in case (b) is $18\text{kg}$ ,net force in case (c) is $\text{4}0\text{N}$ ,mass of the object in case (c) is $30\text{kg}$ ,net force in case (d) is $\text{5}0\text{N}$ ,and mass of the object in case (d) is $32\text{kg}$ .

Write the expression for the force.

$F=ma$

$a=\frac{F}{m}\mathrm{.........}(\text{I})$

Here,

$a$ is the acceleration

$m$ is the mass

Substitute $\text{2}0\text{N}$ for $F$ and $10\text{kg}$ for $m$ in the equation (1) to get $a$ for case a.

$a_a=\frac{20\text{N}}{10\text{kg}}$

$a_a=\frac{20\text{N}}{10\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}$

$a_a=2\text{m}/{\text{s}}^{\text{2}}$

Thus, the acceleration of the object in case (a) is $a_a=2\text{m}/{\text{s}}^{\text{2}}$ .

Substitute $\text{3}0\text{N}$ for $F$ and $18\text{kg}$ for $m$ in the equation (I) to get $a$ for case b.

$a_b=\frac{30\text{N}}{10\text{kg}}$

$a_b=\frac{30\text{N}}{18\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}$

$a_b=1.7\text{m}/{\text{s}}^{\text{2}}$

Thus, the acceleration of the object in case (b) is $a_b=1.7\text{m}/{\text{s}}^{\text{2}}$ .

$\text{4}0\text{N}$ for $F$ and $30\text{kg}$ for $m$ in the equation (I) to get $a$ for case c.

$a_c=\frac{40\text{N}}{30\text{kg}}$

$a_c=\frac{40\text{N}}{30\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}$

$a_c=1.4\text{m}/{\text{s}}^{\text{2}}$

Thus, the acceleration of the object in case (c) is $a_c=1.4\text{m}/{\text{s}}^{\text{2}}$ .

Substitute $\text{5}0\text{N}$ for $F$ and $32\text{kg}$ for $m$ in the equation (I) to get $a$ for case d.

$a_d=\frac{50\text{N}}{32\text{kg}}$

$a_d=\frac{50\text{N}}{32\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}$

$a_d=1.6\text{m}/{\text{s}}^{\text{2}}$

Thus, the acceleration of the object in case (d) is $a_d=1.6\text{m}/{\text{s}}^{\text{2}}$ .

Conclusion:

Therefore, the ranking from greatest to least of acceleration of the objects is $a_a>a_b>a_d>a_c.$