The sled travels a distance of 60.2 m before coming to rest.
The girl reaches the ground with a speed u. She then travels forward on the hard icy snow, which has a coefficient of kinetic friction [tex]\mu _k[/tex] . She comes to rest after travelling a distance s on the ground, due to the friction between the sled and the ground.
If the force of friction is f , the mass of the girl and the sled is m and the acceleration due to gravity is g, then,
[tex]f=\mu_kmg[/tex] .......(1)
This force exerts a decelerating force on the sled. If the deceleration of the sled is a, then,
[tex]f=ma[/tex] .......(2)
From equations (1) and (2),
[tex]f=\mu_kmg=-ma\\ a=-\mu_kg[/tex]......(3)
Substitute 0.038 for [tex]\mu _k[/tex] and 9.81 m/s²for g.
[tex]a=-\mu_kg\\ =-(0.038)(9.81m/s^2)\\ =-0.3728m/s^2[/tex]
Since the girl comes to rest, its final velocity is 0. Substitute 6.7 m/s for u and -0.3728m/s² for a in the equation of motion
[tex]v^2=u^2+2as[/tex]
Solve for s.
[tex]v^2=u^2+2as\\ (0m/s)^2=(6.7m/s)^2+2(-0.3728m/s^2)s\\ s=\frac{(6.7m/s)^2}{2(0.3728m/s^2)} \\ =60.2m[/tex]
Thus, the girl travels a distance of 60.2 m before coming to rest.
The distance traveled by the sled on the level ground before coming to a rest is 60.27 meters.
Given the following data:
To find how far (distance) the sled travel on the level ground before coming to a rest:
Mathematically, the force of kinetic friction is given by the formula;
[tex]F_k = umg[/tex] ....equation 1
Where;
The decelerating force exerted on the sled is given by the formula:
[tex]F = -ma[/tex] ...equation 2.
First of all, we would determine the deceleration of the sled before coming to a rest.
Equating the two equations, we have:
[tex]umg = -ma\\a = -ug[/tex]
Substituting the given parameters into the formula, we have;
[tex]a = -(0.038)(9.8)[/tex]
a = -0.3724 [tex]m/s^2[/tex]
Now, we can determine how far (distance) the sled traveled by using the third equation of motion;
[tex]V^2 = U^2 + 2aS\\\\0^2 = 6.7^2 + 2(-0.3724)S\\\\0 = 44.89 - 0.7448S\\\\0.7448S = 44.89\\\\S = \frac{44.89}{0.7448}[/tex]
Distance, S = 60.27 meters.
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