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

• Nov 1st 2007, 01:08 PM
rcmango
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.
• Nov 1st 2007, 05:34 PM
topsquark
Quote:

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
• Nov 3rd 2007, 06:36 PM
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.
• Nov 4th 2007, 02:20 AM
topsquark
Quote:

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.

Quote:

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