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

**rcmango** · November 1st 2007, 01:08 PM

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 µk = 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.

**topsquark** · November 1st 2007, 05:34 PM

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 µk = 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:
$W_{nc} = W_{friction} = \Delta KE$

-Dan

**rcmango** · November 3rd 2007, 06:36 PM

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.

**topsquark** · November 4th 2007, 02:20 AM

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:
$W_{nc} = W_{friction} = \Delta KE$

-Dan

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

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

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

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

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

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

-Dan

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