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Q104P

Expert-verifiedFound in: Page 148

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A four-person bobsled ${\left(\mathrm{total}\mathrm{mass}=630\mathrm{kg}\right)}$**** comes down a straightaway at the start of a bobsled run. The straightaway is 80.0 m**** long and is inclined at a constant angle of ****${\mathbf{10}}{\mathbf{.}}{\mathbf{2}}{\mathbf{\xb0}}$with the horizontal. Assume that the combined effects of friction and air drag produce on the bobsled a constant force of 62.0 N**** that acts parallel to the incline and up the incline. Answer the following questions to three significant digits.**

**(a) If the speed of the bobsled at the start of the run is 6.20 m/s****, how long does the bobsled take to come down the straightaway?**

**(b) Suppose the crew is able to reduce the effects of friction and air drag to 42.0 N****. For the same initial velocity, how long does the bobsled now take to come down the straightaway?**

$\mathbf{a}\mathbf{)}\mathbf{}6.80\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathbf{b}\mathbf{)}\mathbf{}6.76\mathrm{s}$

$\mathrm{Total}\mathrm{mass}\left(m\right)=630\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Lenght}\left(s\right)=80.0\mathrm{m}\mathrm{and}\mathrm{is}\mathrm{inclined}\mathrm{at}\mathrm{a}\mathrm{constant}\mathrm{angle}\mathrm{of}10.2\xb0\mathrm{with}\mathrm{the}\mathrm{horizontal}.\phantom{\rule{0ex}{0ex}}\mathrm{Force}\left(f\right)=62.0\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{velocity}\mathrm{is},u=6.20\mathrm{m}/\mathrm{s}$

**The problem deals with Newton’s second law of motion. It tells us that the acceleration a of an object is affected by the net force F acting upon the object and the mass of m the object. **

** **

Formula:

F = ma

The component of the weight along the incline (with downhill understood as the positive direction) is,

$mg\mathrm{sin}\theta $

Where m is the mass and is $\theta $ the angle.

We can write the balance force equation as,

$mg\mathrm{sin}\theta -f=ma$

Substitute the values in the above equation, and we get,

$630\times 9.8\times \mathrm{sin}10.2\xb0-62=630\times a\phantom{\rule{0ex}{0ex}}a=1.937\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{a}=1.64\mathrm{m}/{\mathrm{s}}^{2}$

We know that from the equation of motion, we can write the expression for distance as,

$s=ut+1/2a{t}^{2}$

Substitute the values in the above equation, and we get,

$80.0\mathrm{m}=\left(6.20\mathrm{m}/\mathrm{s}\right)\mathrm{t}+\left(1/2\right)\left(1.64\mathrm{m}/{\mathrm{s}}^{2}\right){\mathrm{t}}^{2}$

The above equation is a quadratic equation of t; we can find the root as,

t = 6.80 s

Thus, the bobsled takes 6.80 s to come down straightaway.

$\mathrm{Force}\left(\mathrm{f}\right)=42.0\mathrm{N}$

We can write the balance force equation as,

$mg\mathrm{sin}\theta -f=ma$

Substitute the values in the above equation, and we get,

$630\times 9.8\times \mathrm{sin}10.2\xb0-42=630\times a\phantom{\rule{0ex}{0ex}}a=1.66\mathrm{m}/{\mathrm{s}}^{2}$

We know that from the equation of motion, we can write the expression for distance as,

$s=ut+1/2a{t}^{2}$

Substitute the values in the above equation, and we get,

$80.0\mathrm{m}=\left(6.20\mathrm{m}/\mathrm{s}\right)\mathrm{t}+\left(1/2\right)\left(1.66\mathrm{m}/{\mathrm{s}}^{2}\right){\mathrm{t}}^{2}$

The above equation is a quadratic equation of t; we can find the root as,

t = 6.76 s

Thus, the bobsled takes 6.76 s to come down the straightaway.

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