Physics, 26.08.2019 04:50, sadielebaron

A60.0-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.)

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answered: GracieMaeB

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:

Mass of skier = 60 kgInitial speed = 12.0 m/sHeight = 2.50Coefficient of friction = 0.08

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:

$Distance = \frac{h}{sin \theta} \\\\Distance = \frac{2.5}{sin (35)} \\\\Distance = \frac{2.5}{0.5736}$

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:

K.E_i = K.E_f + P.E + Work

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_f^2 + mgh + Fd$ .....equation 1.

Since we're solving for her final speed, we would rearrange the formula:

$\frac{1}{2} mv_f^2 = \frac{1}{2} mv_i^2 - mgh - Fd$ .....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:

$F = umgcos\theta\\\\F = 0.08(60)(9.8)cos(35)\\\\F = 47.04(0.8192)$

Force, F = 38.54 Newton.

Substituting the values into eqn 2, we have:

$\frac{1}{2} (60)v_f^2 = \frac{1}{2} (60)(12)^2 - 60(9.8)(2.50) - 38.54(4.36)\\\\30v_f^2 = 4320 - 1470 - 168.03\\\\30v_f^2 = 4320 - 1638.03\\\\30v_f^2 = 2681.97\\\\v_f^2 = \frac{2681.97}{30}\\\\v_f^2 = 89.4\\\\v_f = \sqrt{89.4}$

Final speed, v_f = 9.5 m/s

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answered: mdegracia73

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

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

A60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees

answered: Guest

Full change in the electric plus system

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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A60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees...