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Q103P

Expert-verifiedFound in: Page 211

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A sprinter who weighs ****670 N**** runs the first 7.0 m**** of a race in ****1.6 s****, starting from rest and accelerating uniformly. What are the sprinter’s **

**Speed and****Kinetic energy at the end of the****1.6 s****?****What average power does the sprinter generate during the****1.6 s****interval?**

- The speed of the sprinter is 8.8 m/s .
- Sprinter’s kinetic energy is 2.6 kJ .
- Average power is 1.6 kW .

Distance, S = 7.0 m

Time t = 1.6 s

Force, F = 670 N

Initial velocity, ${v}_{0}=0$

**Use the four equations of motion to solve such problems. Use the first and second equations of motion to get the value for acceleration and velocity accordingly. **

**The average power of the sprinter generated during the given time interval can be found by dividing the kinetic energy produced by the sprinter.**

**Formulae:**

Equations of motion,

$S=vt+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}{v}_{f}={v}_{0}+at$

Here, S is the displacement, v is the average velocity, ${v}_{f}$ is the final velocity, ${v}_{0}$ is the initial velocity, a is the acceleration, and t is time.

Kinetic energy is define by,

$KE=\frac{1}{2}m{v}^{2}$

Here, m is the mass.

Average power is given by,

${P}_{avg}=\frac{KE}{t}$

From the equation of motion, you have

$S={v}_{0}t+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}7.0\mathrm{m}=0+0.5\times \mathrm{a}\times {\left(1.6\mathrm{s}\right)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{a}=5.469\mathrm{m}/{\mathrm{s}}^{2}$

Using the second equation of motion,

$\mathrm{v}={\mathrm{v}}_{0}+\mathrm{at}\phantom{\rule{0ex}{0ex}}=0+5.469\mathrm{m}/{\mathrm{s}}^{2}\times 1.6\mathrm{s}\phantom{\rule{0ex}{0ex}}=8.8\mathrm{m}/\mathrm{s}$

Hence, the speed of the sprinter is 8.8 m/s .

You know that kinetic energy is,

$KE=\frac{1}{2}m{v}^{2}$

But from Newton’s second law,

$\mathrm{m}=\mathrm{F}/\mathrm{g}\phantom{\rule{0ex}{0ex}}=\frac{670\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^{2}}\phantom{\rule{0ex}{0ex}}=68.36\mathrm{kg}$

Therefore, the kinetic energy will be,

$\mathrm{KE}=0.5\times \left(68.36\mathrm{kg}\right)\times {\left(8.8\mathrm{m}/\mathrm{s}\right)}^{2}\phantom{\rule{0ex}{0ex}}=2617.18\mathrm{J}\phantom{\rule{0ex}{0ex}}=2.6\mathrm{kJ}$

Hence, the sprinter’s kinetic energy is 2.6 kJ.

Average power can be written as

${P}_{avg}=\frac{KE}{t}\phantom{\rule{0ex}{0ex}}=\frac{2617.18\mathrm{J}}{1.6s}\phantom{\rule{0ex}{0ex}}=1635.74\mathrm{W}\phantom{\rule{0ex}{0ex}}=1.6\mathrm{kW}$

Hence, the average power is 1.6 W.

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