(a) 0.8 m/s^2
The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:
[tex]F=ma[/tex]
where
m = 5.0 kg is the mass of the sled
a is the acceleration of the sled
Solving the equation for a, we find:
[tex]a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2[/tex]
(b) 0.098 m/s^2
According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:
[tex]F=ma[/tex]
where
m = 41 kg is the mass of the girl
a is the acceleration of the girl
Solving the equation for a, we find:
[tex]a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2[/tex]
(c) 5.8 s
Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:
[tex]x(t)=\frac{1}{2}a_g t^2[/tex]
where [tex]a_g = 0.098 m/s^2[/tex] is the acceleration of the girl.
The sled starts instead its motion from x = 15 m, so its position at time t is given by
[tex]x'(t)=15-\frac{1}{2}a_s t^2[/tex]
where [tex]a_s=0.8 m/s^2[/tex] is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.
The girl and the sled meet when x(t) = x'(t). So, we find:
[tex]\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s[/tex]
