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Q61PE

Expert-verifiedFound in: Page 264

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A 75.0-kg cross-country skier is climbing a 3.0ΒΊ slope at a constant speed of 2.00 m/s and encounters air resistance of 25.0 N. Find his power output for work done against the gravitational force and air resistance. **

**(b) What average force does he exert backward on the snow to accomplish this? **

**(c) If he continues to exert this force and to experience the same air resistance when he reaches a level area, how long will it take him to reach a velocity of 10.0 m/s?**

(a) The power output is $127W$ .

(b) The average force is $63.5N$ .

(c) Time taken to reach a velocity of $10m/s$ is $9.4s$.

The power is given as,

$P=\frac{W}{t}$ (1.1)

Here, W stands for completed work, and t stands for time.

The work done is given as,

$W=Fs$ (1.2)

Here, F is the force and s is the distance traveled.

From equation (1.1) and (1.2),

$\begin{array}{rcl}P& =& \frac{Fs}{t}\\ & =& F\frac{s}{t}\\ & =& Fv\\ & & \end{array}$

Since velocity is given as,

$v=\frac{s}{t}$

Free body diagram

Here, m is the mass of the person (m = 75.0 kg), g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$ , $\mathrm{\Xi \u0388}$ is the angle of inclination $\left(\mathrm{\Xi \u0388}=3.0{\rm B}\xb0\right)$ , and f is the air resistance $\left(f=25N\right)$ ,F is the net force, and u is the initial velocity (u = 2 m/s).

(a)

The total force acting on the skier is,

$F=mg\mathrm{sin}\mathrm{\Xi \u0388}+f$

Here, m is the mass of the person (m = 75.0 kg), g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$ , $\mathrm{\Xi \u0388}$ is the angle of inclination $\left(\mathrm{\Xi \u0388}=3.0{\rm B}\xb0\right)$, and f is the air resistance $\left(f=25N\right)$ .

Putting all known values,

$\begin{array}{rcl}F& =& \left(75.0\text{kg}\right)\Gamma \x97\left(9.8\text{m}/{\text{s}}^{2}\right)\Gamma \x97\mathrm{sin}\left(3{\rm B}\xb0\right)+\left(25\text{N}\right)\\ & \beta \x89\x88& 63.5\text{N}\\ & & \end{array}$

The power output is,

$P=Fv$

Putting all known values,

$\begin{array}{rcl}P& =& \left(63.5\text{N}\right)\Gamma \x97\left(2.00\text{m}/\text{s}\right)\\ & =& 127\text{W}\\ & & \end{array}$

Therefore, the required power output is $127W$ .

(b)

The force skier exert backward on snow is,

$F=mg\mathrm{sin}\mathrm{\Xi \u0388}+f$

Here, m is the mass of the person (m = 75.0 kg), g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$, $\mathrm{\Xi \u0388}$is the angle of inclination $\left(\mathrm{\Xi \u0388}=3.0{\rm B}\xb0\right)$, and f is the air resistance $\left(f=25N\right)$ .

Putting all known values,

$\begin{array}{rcl}F& =& \left(75.0\text{kg}\right)\Gamma \x97\left(9.8\text{m}/{\text{s}}^{2}\right)\Gamma \x97\mathrm{sin}\left(3{\rm B}\xb0\right)+\left(25\text{N}\right)\\ & \beta \x89\x88& 63.5\text{N}\\ & & \end{array}$

Therefore, the required average force is $63.5N$ .

(c)

The average force is,

$F=ma$

Here, m is the mass of the person (m = 75.0 kg), and a is the acceleration.

The expression for acceleration is,

$a=\frac{F}{m}$

Putting all known values,

$\begin{array}{rcl}a& =& \frac{63.5\text{N}}{75\text{kg}}\\ & =& 0.85\text{m}/{\text{s}}^{2}\\ & & \end{array}$

The first equation of motion is,

$v=u+at$

Here, v is the final velocity (v = 10 m/s), u is the initial velocity (u = 2 m/s), A is the acceleration of the person $\left(a=0.85\text{m}/{\text{s}}^{2}\right)$ , and T is the time.

The expression for the time is,

$t=\frac{v-u}{a}$

Putting all known values,

$\begin{array}{rcl}t& =& \frac{\left(10\text{m}/\text{s}\right)-\left(2\text{m}/\text{s}\right)}{0.85\text{m}/{\text{s}}^{2}}\\ & =& 9.4\text{s}\\ & & \end{array}$

Therefore, the time taken to reach a velocity of $10m/s$ is $9.4s$ .

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