Answer:
0.8m/s
Explanation:
Weight of mas,F=763 N
Mass of man=[tex]\frac{F}{g}=\frac{763}{9.8}=77.86 kg[/tex]
By using [tex]g=9.8m/s^2[/tex]
Weight of flatcar=F'=3513 N
Mass of flatcar=[tex]\frac{3513}{9.8}=358.5 Kg[/tex]
Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg
Velocity of system=19.8m/s
Let v be the velocity of flatcar with respect to ground
Velocity of man relative to the flatcar=[tex]-4.68m/s[/tex]
Final velocity of man with respect to ground=v-4.68
By using law of conservation of momentum
Initial momentum=Momentum of car+momentum of flatcar
[tex]436.36(19.8)=77.86(v-4.68)+358.5v[/tex]
[tex]8639.928=77.86v-364.3848+358.5v[/tex]
[tex]8639.928+364.3848=436.36 v[/tex]
[tex]9004.3128=436.36v[/tex]
[tex]v=\frac{9004.3128}{436.36}[/tex]
[tex]v=20.6 m/s[/tex]
Initial speed of flatcar=Speed of system
Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s
