Question

Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a distance of 21 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is k0.050. Initially, how fast was the skier going?

Answer 1

The initial velocity of the skier is 4.5 m/s

Explanation:

There is only one force acting on the skier in the horizontal direction, and it is the force of friction, whose magnitude is

$F_f = -\mu_k mg$

where

$\mu_k = 0.050$ is the coefficient of kinetic friction

m is the mass of the skier

$g=9.8 m/s^2$ is the acceleration of gravity

The negative sign is due to the fact that the direction of the force of friction is opposite to the direction of motion of the skier.

According to Newton's second law, the net force acting on the skier is equal to the product between his mass and his acceleration, so we can write:

$F=ma\\ \rightarrow -\mu_k mg = ma$

So, the acceleration of the skier is

$a=-\mu_k g$

Now we can apply the following suvat equation to find the initial velocity of the skier:

$v^2-u^2=2as$

where

v = 0 is the final velocity (he comes to a stop)

u is the initial velocity

s = 21 m is the displacement

a is the acceleration

Substituting the equation for the acceleration and solving for u, we find

$u=\sqrt{v^2-2as}=\sqrt{-2(-\mu_k g) s}=\sqrt{2(0.050)(9.8)(21)}=4.5 m/s$

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