And there it goes...

Splitting up the force of gravity into its components, we get, the force parallel to the slope equal to $mg\sin15$ (call this $F_1$ ) and the force perpendicular to the slope equal to $mg\cos15$ (call this $F_2$ ).

Since the ski is not accelerating, a frictional force equal and opposite to $F_1$ must act on the ski. Also, the normal force exerted by the ground also has to be equal and opposite to $F_2$ .

Therefore, $\mu = \frac{F_f}{F_n} = \frac{mg\sin15}{mg\cos15} = \tan15 = 0.267949192$

Let the mass of the ski be $m$ . The weight $mg$ acts vertically downwards . Resolving $mg$ into perpendicular components , we get $mg\cos\theta$ , perpendicular to the incline in the downwards direction and $mg\sin\theta$ along the incline in the direction of its slipping , where $\theta$ is the inclination of the slope . We also have a normal force which is equal to $mg\cos\theta$ since there is no vertical motion about the incline ,i.e. the block only slides on the incline , and we have the frictional force equal to $\mu mg\cos\theta$ acting on the ski , where $\mu$ is the coefficient of friction between the ski and the slope . Since the ski is moving with a constant velocity , therefore , there is no net acceleration along the incline . Thus , $\mu mg\cos\theta = mg\sin\theta \Rightarrow \mu = \tan\theta$ . Putting $\theta = 15^\circ$ , we get $\mu = 0.268$ .

The component of the force of gravity that is down the incline must equal (so as to balance) the force of friction up the incline. This is because the velocity is a constant, so the acceleration is $0$ , and from Newton's Second Law, the Net Force is $0$ . Since the force of gravity is $mg$ , the component parallel to the incline is $mg \sin \theta$ , where $m$ and $\theta$ are the skier's mass and the incline angle, respectively.

Now, note that the force of friction is $F_f = N \mu$ , where $N$ is the normal force and $\mu$ is the coefficient of friction. Since $N = mg \cos \theta$ , we have $F_f = mg \mu \cos \theta.$ Now, setting the two forces equal, we have $mg \sin \theta = mg \mu \cos \theta \implies \tan \theta = mu,$ so $mu = \tan \theta$ .

In our scenario, $\theta = 15^\circ$ , so $mu = \tan 15^\circ = \boxed {0.268}$ .

The forces on the ski can be resolved into components parallel and perpendicular to the slope. Now, the normal force $N$ balances the resolved weight force $mg\cos15^\circ$ . Also, since the ski is sliding, the friction involved is kinetic friction given by $f_k = \mu N$ . Plugging the value of $N = mg\cos15^\circ$ gives the answer.

The body (here, the ski) is travelling at a constant speed which implies the net acceleration on it is zero.

If we make a free body diagram of the ski, we will will notice that:

1) Mg $\sin \theta$ acts along the incline.

$theta$ = Angle made by the incline with respect to the horizontal.

2) Mg $\cos \theta$ acts perpendicular to the incline.

3) Frictional force = $Coefficient of friction \times Normal force (here, Mg\cos \theta$ )

Acceleration along the incline = 0 (Constant velocity )

Mg $\sin \theta$ = $x$ Mg $\cos \theta$

( $x$ = Coefficient of friction)

Therefore, we get, $x$ = $\tan\theta$ = $\tan\ 15$ = 0.268

You only need the angle when you start skiing down to find the coefficient of friction.

The others are irrelevant.

coefficient of friction $=\tan 15^\circ = 0.268$

After drawing the FBD, you find that the x component is equal to 9.8sin15 and the y component to 9.8cos15.

We know that the coefficient of friction is equal to the Friction Force/Normal Force.

Therefore, the coefficient of friction is 9.8sin15/9.8cos15, which is equal to tan15=0.268

Because there is no acceleration, there is no resultant force acting on it.

Equate the horizontal forces acting on the ski

$N \sin \theta = \mu N \cos \theta$

therefore:

$\mu = \tan \theta$

Sub in $\theta = 15$ therefore:

$\mu = 0.2679$

coefficient of fricion = arctan (angle between Normal reaction and friction) here angle is $15^{\circ}$

tan $15^{\circ} = 0.268$

As ski continues moving with a constant velocity hence all force must be balanced to give net zero force on ski.
Now as ski experiences force of gravity (a component) and friction force along the slope and in opposite direction. Hence they must be balanced.
Therefore
$\mu m g \cos \theta = m g \sin \theta$ or
$\mu = \tan \theta = 0.268$ where $\theta$ is angle of slope = 15 degrees

Hello,

as for this, v = 4 m/s(constant) leads to a = 0 m/s/s,

By Fnett = ma,

as the ski was sliding down,

Wx(force respect to the horizontal) - Fk(Frictional Force) = ma

mgsin(15 degrees) - Uk[mgcos(15 degrees)] = m(0)

mgsin(15 degrees) = Uk[mgcos(15 degrees)]

Uk = sin (15 degrees) / cos *15 degrees)

Uk = Tan (15 degrees) = 0.268

Therefore Uk = 0.268(the friction coefficient)....

Thanks ah....