Question

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0° slope at constant speed, as shown below. the coefficient of friction between the sled and the snow is 0.100. (a) how much work is done by friction as the sled moves 30.0 m along the hill? (b) how much work is done by the rope on the sled in this distance? (c) what is the work done by the gravitational force on the sled? (d) what is the total work done?

Answer 1

A. how much work is done by friction as the sled moves 30.0 m along the hill

We use the formula:

friction work = -µmgdcosΘ 

friction work = -0.100 * 90 kg * 9.8 m/s^2 * 30 m * cos 60

friction work = - 1,323 J

 

B. how much work is done by the rope on the sled in this distance?

We use the formula:

rope work = -mgd(sinΘ - µcosΘ) 
rope work = - 90 kg * 9.8 m/s^2 * 30 m (sin 60 – 0.100 * cos 60)

rope work = 21,592 J

 

C. what is the work done by the gravitational force on the sled?

We use the formula:

gravity work = mgdsinΘ 

gravity work = 90 kg * 9.8 m/s^2 * 30 m * sin 60

gravity work = 22,915 J

 

D. what is the total work done?

We add everything:

total work = - 1,323 J + 21,592 J + 22,915 J

total work = 43,184 J

Answer 2

Part (a): the work done by friction is $\boxed{1324.35\text{ J}}$

Part (b): the work done by rope is $\boxed{21614.07\text{ J}}$.

Part (c): the work done by gravitational force is $\boxed{22938.42\text{ J}}$.

Part (d): total work done is $\boxed{43228.14\text{ J}}$.

Further Explanation:

The rope is doing work against the gravity. The friction always acts against the direction of motion. Therefore, the friction is in direction of gravity.

Given:

Mass of victim is $90\text{ kg}$.

The inclination of plane is $60^\circ$.

The coefficient of plane is $0.1$.

The distance travelled on plane is $30\text{ m}$.

Concept:

Equations for free body diagram of victim:

Force equation:

Normal force on sled is $mgcos\theta$.

Friction force:

$f=\mu N$  

Substitute $mgcos\theta$ for $N$ in above equation.

$f=\mu mgcos\theta$    

Here, $f$ is the friction force, $m$ is the mass of victim, $g$ is the gravitational acceleration, $\mu$ is the coefficient of friction and $\theta$ is the inclination of plane.

The gravitational force on the sled is $mgsin\theta$.

Work equation:

(a)

Work done by friction:

$W_f=-fd$  

Substitute $\mu mgcos\theta$ for $f$ in above equation.

$\boxed{W_f=-\mu mdgcos\theta}$  

Substitute $90\text{ kg}$ for $m$, $30\text{ m}$ for $d$, $9.81\text{ m}/\text{s}^2$ for $g$, $0.1$ for $\mu$ and $60^\circ$ for $\theta$ in above equation.

$\begin{aligned}W_f&=-0.1\times90\times30\times9.81\times cos\theta\\&=-1324.35\text{ J}\end{aligned}$  

Negative sign shows that friction is acting against the motion.

Thus, the work done by friction is $\boxed{1324.35\text{ J}}$

(b)  

Work done by rope:

$\boxed{W_r=mdgsin\theta+W_f}$  

Substitute $90\text{ kg}$ for $m$, $30\text{ m}$ for $d$, $9.81\text{ m}/\text{s}^2$ for $g$, $60^\circ$ for $\theta$ and $-1324.35\text{ J}$ for $W_f$ in above equation.

$\begin{aligned}W_r&=90\times30\times9.81\times sin60^\circ-1324.35\text{ J}\\&=21614.07\text{ J}\end{aligned}$  

Thus, the work done by rope is $\boxed{21614.07\text{ J}}$.

(c)

Work done by gravitational force:

$\boxed{W_w=mdgsin\theta}$  

Substitute $90\text{ kg}$ for $m$, $30\text{ m}$ for $d$, $9.81\text{ m}/\text{s}^2$ for $g$ and $60^\circ$ for $\theta$ in above equation.

$\begin{aligned}W_w&=90\times30\times9.81\times sin60^\circ\\&=22938.42\text{ J}\end{aligned}$  

Thus, the work done by gravitational force is $\boxed{22938.42\text{ J}}$.

(d)

Total work done:

$\boxed{W=W_r+W_w+W_f}$  

Substitute $21614.07\text{ J}$ for $W_r$, $22938.42\text{ J}$ for $W_w$ and $\boxed{-1324.35\text{ J}}$ for $W_f$ in above equation.

$\begin{aligned}W&=21614.07+22938.42-1324.35\text{ J}\\&=43228.14\text{ J}\end{aligned}$

Thus, total work done is $\boxed{43228.14\text{ J}}$

Learn more:

1. Motion on a rough surface: brainly.com/question/7031524

2. Motion under the gravitational force: brainly.com/question/10934170

3. Principle of conservation of momentum: brainly.com/question/9484203

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Ski patrol, rescue, sled, victim, mass, 90.0 kg, 60.0°, slope, coefficient, friction, snow, 0.100, work, 30.0 m, along, hill, distance and gravitational force