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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A 75.0-kg cross-country skier is climbing a 3.0º 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 $9.4s$.

See the step by step solution

## Step 1: Relation between power and velocity:

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,

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

## Step 2: Free body diagram

Free body diagram

Here, m is the mass of the person (m = 75.0 kg), g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$ , $\theta$ is the angle of inclination $\left(\theta =3.0°\right)$ , 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).

## Step 3: Man’s power output for work done against the gravitational force and air resistance

(a)

The total force acting on the skier is,

$F=mg\mathrm{sin}\theta +f$

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

Putting all known values,

$\begin{array}{rcl}F& =& \left(75.0\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×\mathrm{sin}\left(3°\right)+\left(25\text{N}\right)\\ & \approx & 63.5\text{N}\\ & & \end{array}$

The power output is,

$P=Fv$

Putting all known values,

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

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

## Step 4: Average force skier exert backward on snow

(b)

The force skier exert backward on snow is,

$F=mg\mathrm{sin}\theta +f$

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

Putting all known values,

$\begin{array}{rcl}F& =& \left(75.0\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×\mathrm{sin}\left(3°\right)+\left(25\text{N}\right)\\ & \approx & 63.5\text{N}\\ & & \end{array}$

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

## Step 5: Time taken to reach a velocity

(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 $9.4s$ .