Question

FIGURE EX$6.10$shows the velocity graph of a $2.0kg$object as it moves along the x-axis. What is the net force acting on this object at $t=1s$? At $4$s? At $7$s?

Short answer

The net force acting on the object,

(a) At $t=1s$= $6N$,

(b) At role="math" localid="1647761415369" $t=4s$= $3N$,

(c) At$t=7s$=$0N$.

Step-by-step solution

Net force :

The total of all of these forces exerted on an item is the net force.

(a) Explanation : 

Object's mass = $2.0kg$travels along the x-axis A graph of velocity vs. time is shown in Figure 1.

The graph shows the acceleration value at $t=1s$is,

$$\begin{gathered} a=\frac{6-0}{2-0} \\ =3m/s^2 \end{gathered}$$

According to Newton's second law,

$F=ma$

$m$= object's mass,

$a$= object's acceleration,

Substitute $2.0kg$for $m$and $3m/s^2$for $a$in above equation .

$$\begin{gathered} F=\left(2.0kg\right)\left(3m/s^2\right) \\ =6kg·m/s^2\times \frac{1N}{1kg·m/s^2} \\ =6N \end{gathered}$$

As a result, the net force applied to the item at $t=1s$is $6N$.

(b) Explanation :

Object's mass = $2.0kg$moves along x-axis.

The graph shows the acceleration value at $t=4s$is,

$$\begin{gathered} a=\frac{12-6}{6-2} \\ =1.5m/s^2 \end{gathered}$$

Substitute $2.0kg$for $m$and $1.5m/s^2$for $a$in equation.

$$\begin{gathered} F=\left(2.0kg\right)\left(1.5m/s^2\right) \\ =3kg·m/s^2\times \frac{1N}{1kg·m/s^2} \\ =3N \end{gathered}$$

As a result, the net force applied to the item at $t=4s$is $3N$.

(c) Explanation : 

Object's mass = $2.0kg$moves along x-axis.

The graph shows the acceleration value at $t=7s$is,

$$\begin{gathered} a=\frac{12-12}{8-6} \\ =0m/s^2 \end{gathered}$$

Substitute $2.0kg$for $m$and $0m/s^2$for $a$in equation.

$$\begin{gathered} F=\left(2.0kg\right)\left(0m/s^2\right) \\ =0N \end{gathered}$$

As a result, the net force applied to the item at $t=7s$is $0N$.