Chapter 3, Problem 2TC

Boxes of chocolates of various masses on a friction-free level table.

Compare from greatest to least:

(a) The net forces on the boxes.

Greatest ___ ___ ___ ___ Least

(b) The acceleration of the boxes.

Greatest ___ ___ ___ ___ Least.

To determine

The comparison between net forces on the boxes$D>A=B=C$ .

Answer to Problem 2TC

The comparison between net forces on the boxes is .

Explanation of Solution

Write the expression for the net forces, when forces acts in opposite direction.

$F=F_1-F_2$ ……(1)

Here,

$F_1$ is the first force.

$F_2$ is the second force.

Perform calculation for case A,

Substitute $10\text{N}$ for $F_1$ and $5\text{N}$ for $F_2$ in equation (1) to get $F$ .

$\begin{array}{c}{F}_A=F_1-F_2\\ F_A=10\text{N}-5\text{N}\\ F_A=5\text{N}\end{array}$

Thus, the net force in case A is $F_A=5\text{N}$ .

Perform calculation for case B,

Substitute $15\text{N}$ for $F_1$ and $10\text{N}$ for $F_2$ in equation (1) to get $F$ .

$\begin{array}{c}{F}_B=F_1-F_2\\ F_B=15\text{N}-10\text{N}\\ F_B=5\text{N}\end{array}$

Thus, the net force in case B is $F_B=5\text{N}$ .

Perform calculation for case C,

Substitute $15\text{N}$ for $F_1$ and $10\text{N}$ for $F_2$ in equation (1) to get,

$\begin{array}{c}{F}_C=F_1-F_2\\ F_C=15\text{N}-10\text{N}\\ F_C=5\text{N}\end{array}$

Thus, the net force in case C is $F_C=5\text{N}$ .

Perform calculation for case D,

Substitute $15\text{N}$ for $F_1$ and $10\text{N}$ for $F_2$ in equation (1),

$\begin{array}{c}{F}_D=F_1-F_2\\ F_D=15\text{N}-5\text{N}\\ F_D=10\text{N}\end{array}$

Thus, the net force in case D is $F_D=10\text{N}$ .

Conclusion:

Therefore, the comparison between net forces on the boxes is $D>A=B=C$ .