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Expert-verified Found in: Page 171 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A luge and its rider, with a total mass of 85 kg, emerge from a downhill track onto a horizontal straight track with an initial speed of 37 m/s. If a force slows them to a stop at a constant rate of 2.0 m/s2, (a) what magnitude F is required for the force, (b) what distance d do they travel while slowing, and (c) what work W is done on them by the force? What are (d) f , (e) d , and (f) W if they, instead, slow at 4.0 m/s2 ?

1. The magnitude of Force when $\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^{2}$is $1.7×{10}^{2}\mathrm{N}$.
2. Distance traveled before coming to a stop when $\mathrm{a}=2.0\frac{\mathrm{m}}{{\mathrm{s}}^{2}}$ is $3.4×{10}^{2}\mathrm{m}$.
3. Work done by the force on the rider when $\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^{2}$ is $-5.8×{10}^{4}\mathrm{J}$.
4. The magnitude of force when $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^{2}$ is $3.4×{10}^{2}\mathrm{N}$.
5. Distance traveled before coming to a stop when $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^{2}$ is $1.7×{10}^{2}\mathrm{m}$.
6. Work done by the force on the rider when $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^{2}$ is $-5.8×{10}^{4}\mathrm{J}$.
See the step by step solution

## Step 1: Given data

i) Mass of the rider is, m=85 kg.

ii) Initial velocity is, ${\mathrm{v}}_{\mathrm{i}}=37\frac{\mathrm{m}}{\mathrm{s}}$.

iii) Acceleration is, ${\mathrm{a}}_{1}=2.0\frac{\mathrm{m}}{{\mathrm{s}}^{2}}$.

## Step 2: Understanding the concept

As we are given the acceleration and mass of the rider, using Newton’s 2nd law, we can find the force acting on them. As we know the initial velocity, final velocity, and acceleration, we can find the distance traveled. As we get the force and displacement, we can find the work done by the force. Using the same procedure, we can find the force, distance, and work done when the rider has an acceleration of ${\mathbf{a}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}\frac{\mathbf{m}}{{\mathbf{s}}^{\mathbf{2}}}$.

## Step 3: (a) Calculate the magnitude of F required if it slows them to a stop at a constant rate of  2.0 m/s2

The relationship between force exerted on a particle mass m of the particle and acceleration a of the particle can be written as,

$\mathrm{F}=\mathrm{ma}$ (1)

Substitute the values in the given expression, and we get the magnitude of the force as,

$\left|\stackrel{\to }{\mathrm{F}}\right|=\left|85×\left(-2.0\right)\right|\phantom{\rule{0ex}{0ex}}\mathrm{F}=\left|-1.7×{10}^{2}\mathrm{N}\right|\phantom{\rule{0ex}{0ex}}\mathrm{F}=1.7×{10}^{2}\mathrm{N}$

Thus, the magnitude of Force when $\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^{2}$ is $1.7×{10}^{2}\mathrm{N}$.

## Step 4: (b) Calculate the distance d that they travel while slowing down

Newton's law of motion, we can write the below expression and calculate the traveled distance x as,

${\mathrm{v}}_{\mathrm{f}}^{2}={\mathrm{v}}_{\mathrm{i}}^{2}+2\mathrm{ax}$ (2)

As the rider is coming to a stop, so we can say that the final velocity is,${\mathrm{v}}_{\mathrm{f}}=0\frac{\mathrm{m}}{\mathrm{s}}$

Substitute the values in the given expression, and we get,

$\begin{array}{rcl}0& =& {37}^{2}-\left(2×2×\mathrm{d}\right)\\ {37}^{2}& =& 4\mathrm{d}\\ \mathrm{d}& =& \frac{1369}{4}\\ \mathrm{d}& =& 3.4×{10}^{2}\mathrm{m}\end{array}$

Thus, the distance traveled before coming to a stop when $\mathrm{a}=2.0\frac{\mathrm{m}}{{\mathrm{s}}^{2}}$ is $3.4×{10}^{2}\mathrm{m}$.

## Step 5: (c) Calculate work done on them by the force

If the force F is exerted on a body and it traveled distance d, then the expression for the work done can be expressed as,

$\mathrm{W}=\mathrm{Fd}$ (3)

Substitute the values in the given expression, and we get,

$\mathrm{W}=\left(-1.7×{10}^{2}\right)×\left(3.4×{10}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{W}=-5.8×{10}^{4}\mathrm{J}$

Thus, work done by the force on the rider when $\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^{2}$ is $-5.8×{10}^{4}\mathrm{J}$.

## Step 6: (d) Calculate the magnitude of F required if it slows them to a stop at a constant rate of  4.0 m/s2

Given that,

$\mathrm{a}=4.0\frac{\mathrm{m}}{{\mathrm{s}}^{2}}$

From equation 1, we can calculate the magnitude of the force as,

$\left|\stackrel{\to }{\mathrm{F}}\right|=\left|85×\left(-4.0\right)\right|\phantom{\rule{0ex}{0ex}}\mathrm{F}=\left|\left(-3.4×{10}^{2}\mathrm{N}\right)\right|\phantom{\rule{0ex}{0ex}}\mathrm{F}=3.4×{10}^{2}\mathrm{N}$

Thus, the magnitude of force when data-custom-editor="chemistry" $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^{2}$ is $3.4×{10}^{2}\mathrm{N}$.

## Step 7: (e) Calculate the distance d that they travel while slowing down

Given that,

$\mathrm{a}=4.0\frac{\mathrm{m}}{{\mathrm{s}}^{2}}$

From equation 2, we can calculate the distance in this case as,

$\begin{array}{rcl}0& =& {37}^{2}-\left(2×4×\mathrm{d}\right)\\ {37}^{2}& =& 8\mathrm{d}\\ \mathrm{d}& =& \frac{1369}{8}\\ \mathrm{d}& =& 1.7×{10}^{2}\mathrm{m}\end{array}$

Thus, the distance traveled before coming to a stop when $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^{2}$ is $1.7×{10}^{2}\mathrm{m}$

## Step 8: (f) Calculate work done on them by the force

From equation 3, we can calculate the work done as,

$\mathrm{W}=\left(-3.4×{10}^{2}\right)×\left(1.7×{10}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{W}=-5.8×{10}^{4}\mathrm{J}$

Thus, work done by the force on the rider when data-custom-editor="chemistry" $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^{2}$ is $-5.8×{10}^{4}\mathrm{J}$.

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