Question

A truck driver slams on the brakes and the momentum of the truck changes from$\mathbf{<}\mathbf{65}\mathbf{,}\mathbf{000}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{to}\mathbf{<}\mathbf{26}\mathbf{,}\mathbf{000}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}$
$\mathbf{in}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{s}\mathbf{}\mathbf{}$due to a constant force of the road on the wheels of the truck. As a vector, write the net force exerted on the truck by the surroundings.

Short answer

The net force exerted on the truck by the surrounding is$<-9512.2,0,0>\mathrm{N}$.

Step-by-step solution

Definition and the formula of the net force

The rate of change of a particular system’s momentum will be equated to the net force applied to that system.

If the system's momentum changes from initial momentum $\overrightarrow{{\mathrm{p}}_1}$ to final momentum $\overrightarrow{p_2}$over time $\mathbf{∆}\mathbf{T}$, the force acting on the system will be,

localid="1653963994357">→Fnet=p2→-p→1∆t

Finding the net force on the system.

Substitute $<65,000,0,0>\mathrm{kg}.\mathrm{m}/\mathrm{s}$for $\underset{p_1}{\to }$, $<26,000,0,0>\mathrm{kg}.\mathrm{m}/\mathrm{s}$for $\underset{p_2}{\to }$, and $4.1s$for $∆\mathrm{t}$into the formula of the net force acting on the system.

role="math" localid="1653965078905" $$\begin{gathered} \overrightarrow{{\mathbf{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}}\mathbf{}\mathbf{=}\frac{\mathbf{<}\mathbf{26}\mathbf{,}\mathbf{000}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{kg}\mathbf{.}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{<}\mathbf{65}\mathbf{,}\mathbf{000}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{kg}\mathbf{.}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{4}\mathbf{.}\mathbf{1}\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{<}\mathbf{26}\mathbf{,}\mathbf{000}\mathbf{-}\mathbf{65}\mathbf{,}\mathbf{000}\mathbf{,}\mathbf{0}\mathbf{-}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{-}\mathbf{0}\mathbf{>}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{4}\mathbf{.}\mathbf{1}\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{<}\mathbf{-}\mathbf{39}\mathbf{,}\mathbf{000}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{4}\mathbf{.}\mathbf{1}\mathbf{s}}\mathbf{}\mathbf{.}\mathbf{\left(}\begin{array}{c}\mathbf{1}\mathbf{N}\\ \mathbf{1}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}\end{array}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{<}\mathbf{-}\mathbf{9512}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{N} \\ \mathbf{Therefore}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{net}\mathbf{}\mathbf{force}\mathbf{}\mathbf{on}\mathbf{}\mathbf{the}\mathbf{}\mathbf{system}\mathbf{}\mathbf{is}\mathbf{}\mathbf{}\mathbf{<}\mathbf{-}\mathbf{9512}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{N}\mathbf{.} \end{gathered}$$