Answer:
The force is 272.73 newtons
Explanation:
We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:
[tex]\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}[/tex] (1)
Impulse is also defined as average force times the time the force is applied:
[tex]\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) [/tex] (2)
By (2) on (1):
[tex]\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i} [/tex]
solving for [tex] \overrightarrow{F}_{avg}[/tex]:
[tex]\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t} [/tex] (3)
We already know Δt is equal to 0.22 s, all we should do now is to find [tex] \overrightarrow{p}_{f}-\overrightarrow{p}_{i}[/tex] and put on (3) ([tex] \overrightarrow{p_{i}}[/tex] the initial momentum and [tex]\overrightarrow{p_{f}}[/tex] the final momentum). Linear momentum is defined as [tex]\overrightarrow{p}=m\overrightarrow{v}[/tex] , using that on (3):
[tex]\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}[/tex] (4)
Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left [tex]\overrightarrow{v_{i}}=+3\, \frac{m}{s}[/tex] and [tex]\overrightarrow{v_{f}}=-3\, \frac{m}{s}[/tex] so (4) becomes:
[tex]\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})[/tex]
[tex]\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s} [/tex] (5)
Using (5) on (3):
[tex] \overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s} [/tex]
[tex] F_{avg}=272.73N[/tex]
