Answer:v=3.08 m/s
Explanation:
Given
mass of student [tex]m=21 kg[/tex]
distance moved [tex]d=10 m[/tex]
Force applied [tex]F=10 N[/tex]
acceleration of system during application of force is a
[tex]a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2[/tex]
using [tex]v^2-u^2=2 as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex]v^2-0=2\times 0.476\times 10[/tex]
[tex]v=\sqrt{9.52}[/tex]
[tex]v=3.08 m/s[/tex]
