Question

A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

Answer 1

Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.= 
Wt. of skier. 

Fp=588*sin35 = 337 N.=Force parallel to 
incline. 
Fv = 588*cos35 = 482 N. = Force perpendicular to incline. 

Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force 
of kinetic friction. 
d =h/sinA = 2.5/sin35 = 4.36 m. 

Ek + Ep = Ekmax - Fk*d 
Ek = Ekmax-Ep-Fk*d 
Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. 
Ek = 0.5m*V^2 = 2682 J. 
30*V^2 = 2682 
V^2 = 89.4 
V = 9.5 m/s = Final velocity.

Answer 2

Answer:

9.46 m/s

Explanation:

Let's start by writing the equation of the forces along the two directions:

- Parallel to the ramp: $-mg sin \theta - \mu N = ma$

where the first term is the component of the weight parallel to the ramp, and the second term is the frictional force

- Perpendicular to the ramp: $N-mg cos \theta =0$

where N is the normal reaction of the ramp and the second term is the component of the weight perpendicular to the ramp

Solving the second equation, we get

$N=mg cos \theta$

And we can substitute it into the first equation:

$-mg sin \theta - \mu mg cos \theta = ma$

From this equation, we can find the acceleration of the skier:

$a=g sin \theta - \mu g cos \theta = -(9.8 m/s^2)(sin 35^{\circ})-(0.08)(9.8 m/s^2)(cos 35^{\circ})=-6.26 m/s^2$

And the sign is negative because the acceleration is downward along the ramp.

By using trigonometry, we can also find the length of the ramp: in fact, the height is h=2.50 m, while the angle is $\theta=35^{\circ}$, so the length is given by

$L=\frac{h}{sin \theta}=\frac{2.50 m}{sin 35^{\circ}}=4.36 m$

And now we can find the final speed of the skier at the top by using the following SUVAT equation:

$v^2 - u^2 = 2aL$

where v = ? is the final speed and u = 12.0 m/s is the initial speed. Substituting, we find

$v=\sqrt{u^2+2aL}=\sqrt{(12 m/s)^2+2(-6.26 m/s^2)(4.36 m)}=9.46 m/s$