At \(t = 16.0\;s\)an object with mass \(4\;kg\)was observed to have a velocity of \(\left\langle {9,29, - 10} \right\rangle \;m/s\) at \(t = 16.2\;s\left\langle {180, - 180,700} \right\rangle \;N\)\(\)its velocity was \(\left\langle {18,20,25} \right\rangle \;m/s\) what was the average net force acting on the object?

At \(t = 16.0\;s\) an object with mass \(4\;kg\) was observed to have a velocity of

\(\left\langle {9,29, - 10} \right\rangle \;m/s\)at\(t = 16.2\;s\left\langle {180, - 180,700} \right\rangle \;N\)\(\)its velocity was\(\left\langle {18,20,25} \right\rangle \;m/s\)

The rate of change in momentum of an object is the net force acting on it.

The relationship between the object's initial momentum\({\vec p_1}\), end momentum\({\vec p_2}\), and force \(({\vec F_{act }})\)

\({\vec F_{net}} = \frac{{{{\vec p}_2} - {{\vec p}_1}}}{{\Delta t}}\)

If\(m\)is the object's mass and\({\vec v_1}\)is the object's beginning velocity, the starting momentum is\({\vec p_1} = m{\vec v_1}\)

Let\({\vec v_2}\)be the object's final velocity, and then the final momentum is,

\({\vec p_2} = m{\vec v_2}\)

Now insert\(m{\vec v_2}\)for\({\vec p_2}\), and\(m{\vec v_1}\)for\({\vec p_1}\).

\(\begin{aligned}{l}{{\vec F}_{net }} = \frac{{m{{\vec v}_2} - m{{\vec v}_1}}}{{\Delta t}}\\{{\vec F}_{net }} = \frac{{m\left( {{{\vec v}_2} - {{\vec v}_1}} \right)}}{{\Delta t}}\end{aligned}\) …… (1)

The initial velocity is\(16.0\;{\rm{s}}\), while the final velocity is\(16.2\;{\rm{s}}\).

The change in time is calculated as shown below,

\(\begin{aligned}{c}\Delta t = 16.2\;{\rm{s}} - 16.0\;{\rm{s}}\\ = 0.2\;{\rm{s}}\end{aligned}\)

Substitute\(\left\langle {9,29, - 10} \right\rangle \;{\rm{m/s}}\)for\({\vec v_1}\),\(\left\langle {18,20,25} \right\rangle {\rm{ m/s}}\)for\({\vec v_2}\),and\(4\;{\rm{kg}}\)for\(m\)in the equation 1

\(\begin{aligned}{c}{{\vec F}_{{\rm{net }}}} = \frac{{(4\;)(\left\langle {18,20,25} \right\rangle \; - \left\langle {9,29, - 10} \right\rangle )}}{{0.2\;}}\\ = \frac{{(4\;)(\left\langle {18 - 9,20 - 29,25 + 10} \right\rangle \;)}}{{0.2\;}}\\ = \frac{{(4\;)(\left\langle {9, - 9,35} \right\rangle \;)}}{{0.2\;{\rm{s}}}}\\ = \left\langle {180, - 180,700} \right\rangle \;\;{\rm{N}}\end{aligned}\)

Thus, the net force is\(\left\langle {180, - 180,700} \right\rangle \;\;{\rm{N}}\).