Question

A girl on a sled coasts down a hill. Her speed is 6.0 m/s when she reaches level ground at the bottom. The coefficient of kinetic friction between the sled's runners and the hard, icy snow is 0.052, and the girl and sled together weigh 688 N. How far does the sled travel on the level ground before coming to rest?

Answer #1

Ans:-

Given data:- vi = 6m/s μk = 0.052, w = 688N, x = ?

We have to calculate x = ?

Before that we need acceleration so first find out force

Fk = μk*N

N = mg = 688N

Fk = 0.052*688 = 35.78N

Now

F = mg

m= F/g = 688/9.8 = 70.20kg

F = ma

Here force is frictional force

a = F/m =35.78/70.20 =0.510m/s^2

use kinematics equation

V^2 = V0^2 + 2a(∆x)

Here final velocity is zero because we have to calculate distance where she is stops

0 = 6^2 + 2*(-0.510)*x

Here she is de-accelerate so acceleration is -ve

X= 1.02/36

X= 0.028m

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