Soccer !

We know: $F=ma$ → $a=\frac{F}{m}$

And: $v^{2}-v_{0}^{2} = 2as$

Thus, $v^{2} = 2\frac{F}{m}s → s=\frac{v^{2}m}{2F}$

So, The soccer ball travel $s=\frac{5^{2}\times2}{2\times 50} = 0.5$ m

given that F=50n,M=2kg v=5m/s S=? as it was stationary U=O WE KNOW f=ma therfor A=F/M=50/2 =25 WE KNOW (V^2-U^2)/2A =S,, 5^2/2into25 =0.5m

given that, F=50n, m=2kg V=5m/s AND S=? as ball was stationary so , u=0m/s WE know, F=ma a=f/m=50/2=25 t=(v-u)/a=(5-0)/25=1/5 S=ut+.5at^2=0+.5 25 1/.25 =.5m

d=0.5

$W = \triangle Ek$

$F x s = \frac{1}{2} \times m \times (V_t^{2} - V_0^{2})$

$50s = \frac{1}{2} \times 2 \times (5^{2} - 0^{2})$

$s = \frac{1\times 25}{50} = 0.5$