Physics, 04.07.2019 01:00 haileyhale5

Agirl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 6.7 m/s. the coefficient of kinetic friction between the sled’s runners and the hard, icy snow is 0.038, and the girl and sled together weigh 603 n. the acceleration of gravity is 9.81 m/s 2 . how far does the sled travel on the level ground before coming to a rest?

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Answer from: melindapeckens

The distance traveled by the sled on the level ground before coming to a rest is 60.27 meters.

Given the following data:

Final speed = 6.7 m/sCoefficient of kinetic friction = 0.038Weight of girl and sled together = 603 NewtonAcceleration of gravity = 9.81 m/s^2

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;

F_k = umg ....equation 1

Where;

Fk represents the force of kinetic friction. μ represents the coefficient of friction.m represents the mass.g is the acceleration due to gravity.

The decelerating force exerted on the sled is given by the formula:

F = -ma ...equation 2.

First of all, we would determine the deceleration of the sled before coming to a rest.

Equating the two equations, we have:

$umg = -ma\\a = -ug$

Substituting the given parameters into the formula, we have;

a = -(0.038)(9.8)

a = -0.3724 m/s^2

Now, we can determine how far (distance) the sled traveled by using the third equation of motion;

$V^2 = U^2 + 2aS\\\\0^2 = 6.7^2 + 2(-0.3724)S\\\\0 = 44.89 - 0.7448S\\\\0.7448S = 44.89\\\\S = \frac{44.89}{0.7448}$

Distance, S = 60.27 meters.

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Answer from: constipatedcow18

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 $\mu _k$ . 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,

$f=\mu_kmg$ .......(1)

This force exerts a decelerating force on the sled. If the deceleration of the sled is a, then,

f=ma .......(2)

From equations (1) and (2),

$f=\mu_kmg=-ma\\ a=-\mu_kg$ ......(3)

Substitute 0.038 for $\mu _k$ and 9.81 m/s²for g.

$a=-\mu_kg\\ =-(0.038)(9.81m/s^2)\\ =-0.3728m/s^2$

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

v^2=u^2+2as

Solve for s.

$v^2=u^2+2as\\ (0m/s)^2=(6.7m/s)^2+2(-0.3728m/s^2)s\\ s=\frac{(6.7m/s)^2}{2(0.3728m/s^2)} \\ =60.2m$

Thus, the girl travels a distance of 60.2 m before coming to rest.

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Agirl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 6.7 m/s. the...

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