Physics Laws of Motion

As per Newton third law, the horse exerts a force on the ground, the ground will also exert a force on the horse.

I = m (2v) = 1 * 2 * 15 = 30 kg m s^–1

From F.B. Ds of

$9kg\to 90-T_1=9a\dots .\left(1\right)$

$6kg\to T_1-T_2=6a\dots ..\left(2\right)$

$5kg\to T_2-50=5a\dots ..\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)$

$40=20a?a=2m/s^2$

Balancing torque about COM

$T_A\left(\frac{l}{2}\right)=T_B\left(\frac{l}{6}\right)$

$?\frac{T_A}{T_B}=\frac{1}{3}$

From equation of uniformly accelerated motion

$v=u+at$

$7.5=15+5a$

$a=-1.5m/s^2$

$nowF=ma$

$\left|\stackrel{?}{F}\right|=120*1.5$

$=180N$

From F.B. Ds of

$9kg\to 90-T_1=9a\dots .\left(1\right)$

$6kg\to T_1-T_2=6a\dots ..\left(2\right)$

$5kg\to T_2-50=5a\dots ..\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)$

$40=20a?a=2m/s^2$

Rate of burning of fuel $\left(\frac{dm}{dt}\right)=1kg/s$  and velocity of ejected fuel

$\left(v\right)=60km/s$

$=60*10^{3}m/s.$  

Forcev =  Rate of change of momentum

$=\frac{dp}{dt}=\frac{d\left(m\nu \right)}{dt}=v\frac{dm}{dt}=\left(60*10^3\right)*1$

  $=60000$ N.

Force $\left(F\right)=10N$  and acceleration