Math

QuestionSuppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg , down a 6060^{\circ} slope at constant speed, as shown in the Figure. The coefficient of friction between the sled and the snow is 0.100 . (b) How much work is done by the rope on the sled in this distance? Give your answer in kJ

Studdy Solution

STEP 1

What is this asking? We need to figure out how much work the rope does on the sled as it slides down a slope at a constant speed. Watch out! Constant speed means *zero acceleration*!
Don't forget to consider all the forces acting on the sled.

STEP 2

1. Analyze forces
2. Calculate tension
3. Calculate work

STEP 3

Alright, let's break this down!
Since the sled is moving at a constant speed, the *net force* acting on it is **zero**.
This means all forces balance each other out!
We've got **three** main forces in play: **gravity**, **friction**, and the **tension** from the rope.

STEP 4

Gravity pulls straight down, but we're only interested in the component of gravity *along* the slope.
This is given by mgsin(θ)mg\sin(\theta), where mm is the **mass** (90.0 kg90.0 \text{ kg}), gg is the **acceleration due to gravity** (9.8 m/s29.8 \text{ m/s}^2), and θ\theta is the **angle of the slope** (6060^\circ).

STEP 5

Next up is friction!
It acts *opposite* the direction of motion, up the slope.
The force of friction is given by μN\mu N, where μ\mu is the **coefficient of friction** (0.1000.100) and NN is the **normal force**.
The normal force is equal to the component of gravity *perpendicular* to the slope, which is mgcos(θ)mg\cos(\theta).

STEP 6

Finally, we have the tension in the rope, which also acts up the slope.
Since the net force is zero, the tension and friction forces *together* must balance the component of gravity along the slope.

STEP 7

Let's put it all together mathematically!
We know that the sum of the forces along the slope is zero.
So, we can write: mgsin(θ)Tμmgcos(θ)=0mg\sin(\theta) - T - \mu mg\cos(\theta) = 0, where TT is the **tension** in the rope.

STEP 8

Now, let's plug in the values: 90.0 kg9.8 m/s2sin(60)T0.10090.0 kg9.8 m/s2cos(60)=090.0 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot \sin(60^\circ) - T - 0.100 \cdot 90.0 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot \cos(60^\circ) = 0.

STEP 9

Calculating the terms, we get: 763.8 NT44.1 N=0763.8 \text{ N} - T - 44.1 \text{ N} = 0.

STEP 10

Solving for TT, we find T=763.8 N44.1 N=719.7 NT = 763.8 \text{ N} - 44.1 \text{ N} = \mathbf{719.7 \text{ N}}.
This is the **magnitude** of the tension force.

STEP 11

Work done by a force is given by W=Fdcos(ϕ)W = Fd\cos(\phi), where FF is the **magnitude of the force**, dd is the **distance** over which the force acts, and ϕ\phi is the **angle** between the force and the direction of displacement.

STEP 12

In our case, the rope's tension acts *up* the slope, while the sled moves *down* the slope.
This means the angle between the tension and displacement is 180\mathbf{180^\circ}.
The problem states the sled moves a distance down the slope, let's call it dd.

STEP 13

The work done by the rope is then W=Tdcos(180)W = Td\cos(180^\circ).
Since cos(180)=1\cos(180^\circ) = -1, we have W=TdW = -Td.

STEP 14

We don't have the distance dd given, so we can't calculate a numerical value for the work done by the rope.
However, we can express it in terms of dd: W=719.7d JW = -719.7d \text{ J}.
To convert to kJ, we divide by 1000: W=0.7197d kJW = -0.7197d \text{ kJ}.

STEP 15

The work done by the rope on the sled is 0.7197d-0.7197d kJ, where dd is the distance the sled travels down the slope.
The negative sign indicates the rope is doing *negative* work, meaning it's taking energy *away* from the sled.

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