# Thread: how fast is the skier going, using kinetic friction and length.

1. ## how fast is the skier going, using kinetic friction and length.

A skier slides horizontally along the snow for a distance of 27 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is ľk = 0.050 Initially, how fast was the skier going?
m/s

not sure how to reach the answer with so few variables involved.

equations i thought i might use were: vf = sqrt((2 * (KE) / m)

and KE = 1/2*(m * v^2)

thanks for any extra help.

2. Originally Posted by rcmango
A skier slides horizontally along the snow for a distance of 27 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is ľk = 0.050 Initially, how fast was the skier going?
m/s

not sure how to reach the answer with so few variables involved.

equations i thought i might use were: vf = sqrt((2 * (KE) / m)

and KE = 1/2*(m * v^2)

thanks for any extra help.
Both of these questions are the same, just in a different form.

Consider the following equation:
$\displaystyle W_{nc} = W_{friction} = \Delta KE$

-Dan

3. i've tried using both of these equations to reach the answer which is: 5.14 m/s, however, i can't seem to use the formulas correctly.

what do you mean by chng in KE?

please break this down for me.

thankyou.

4. Originally Posted by rcmango
i've tried using both of these equations to reach the answer which is: 5.14 m/s, however, i can't seem to use the formulas correctly.

what do you mean by chng in KE?

please break this down for me.

thankyou.
Originally Posted by topsquark
Both of these questions are the same, just in a different form.

Consider the following equation:
$\displaystyle W_{nc} = W_{friction} = \Delta KE$

-Dan
The work done by friction is
$\displaystyle W_{friction} = \vec{f_k} \cdot \vec{s}$

Since the friction force is in the opposite direction to the displacement we have
$\displaystyle W_{friction} = \vec{f_k} \cdot \vec{s} = -f_ks$

Now $\displaystyle f_k = \mu_k N$
So what's the normal force? Again we solve this using a Free-Body Diagram. In this case the answer is simple: $\displaystyle N = w = mg$, thus
$\displaystyle W_{friction} = -f_ks = -\mu_k mgs$

Now, the skier starts with a speed $\displaystyle v_0$. At the end, the skier is at rest, so v = 0 m/s. Thus
$\displaystyle \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2$

So finally we have:
$\displaystyle W_{friction} = \Delta KE$

$\displaystyle -\mu_k mgs = -\frac{1}{2}mv_0^2$
which you can solve for $\displaystyle v_0$.

-Dan