Answer:
9.46 m/s
Explanation:
Let's start by writing the equation of the forces along the two directions:
- Parallel to the ramp: [tex]-mg sin \theta - \mu N = ma[/tex]
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: [tex]N-mg cos \theta =0[/tex]
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
[tex]N=mg cos \theta[/tex]
And we can substitute it into the first equation:
[tex]-mg sin \theta - \mu mg cos \theta = ma[/tex]
From this equation, we can find the acceleration of the skier:
[tex]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[/tex]
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 [tex]\theta=35^{\circ}[/tex], so the length is given by
[tex]L=\frac{h}{sin \theta}=\frac{2.50 m}{sin 35^{\circ}}=4.36 m[/tex]
And now we can find the final speed of the skier at the top by using the following SUVAT equation:
[tex]v^2 - u^2 = 2aL[/tex]
where v = ? is the final speed and u = 12.0 m/s is the initial speed. Substituting, we find
[tex]v=\sqrt{u^2+2aL}=\sqrt{(12 m/s)^2+2(-6.26 m/s^2)(4.36 m)}=9.46 m/s[/tex]
Assuming a straight-line path as shown in the figure, the skier's final speed at the top is 9.5 m/s.
Given the following data:
To find the skier's final speed at the top:
First of all, we would determine the distance traveled up the incline (assuming a straight-line path) by using trigonometry:
From Sine law:
[tex]Distance = \frac{h}{sin \theta} \\\\Distance = \frac{2.5}{sin (35)} \\\\Distance = \frac{2.5}{0.5736}[/tex]
Distance, d = 4.36 meters.
The initial kinetic energy possessed by the skier would generate a final kinetic energy, potential energy and heat due to friction as work is done.
Mathematically, this given by the formula:
[tex]K.E_i = K.E_f + P.E + Work[/tex]
[tex]\frac{1}{2} mv_i^2 = \frac{1}{2} mv_f^2 + mgh + Fd[/tex] .....equation 1.
Since we're solving for her final speed, we would rearrange the formula:
[tex]\frac{1}{2} mv_f^2 = \frac{1}{2} mv_i^2 - mgh - Fd[/tex] .....equation 2.
We would apply Newton's Second Law of Motion to determine the force acting on the skier as she ascend straight-line path:
[tex]F = umgcos\theta\\\\F = 0.08(60)(9.8)cos(35)\\\\F = 47.04(0.8192)[/tex]
Force, F = 38.54 Newton.
Substituting the values into eqn 2, we have:
[tex]\frac{1}{2} (60)v_f^2 = \frac{1}{2} (60)(12)^2 - 60(9.8)(2.50) - 38.54(4.36)\\\\30v_f^2 = 4320 - 1470 - 168.03\\\\30v_f^2 = 4320 - 1638.03\\\\30v_f^2 = 2681.97\\\\v_f^2 = \frac{2681.97}{30}\\\\v_f^2 = 89.4\\\\v_f = \sqrt{89.4}[/tex]
Final speed, [tex]v_f[/tex] = 9.5 m/s
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