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A cruise ship with a mass of 1.00 \times 10^7 \,\mathrm{kg}1.00×10 7 kg strikes a pier at a speed of 0.750\,\mathrm{m/s}0.750m/s. It comes to rest after traveling 6.00\,\mathrm{m}6.00m, damaging the ship, the pier, and the tugboat captain’s finances. Calculate the average force exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest, assuming a constant force.)

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davidlcastiblanco

Answer:

[tex]F_i=4.6875x10^5 N[/tex]

Explanation:

Given:

[tex]m=1x10^7 kg[/tex]

[tex]u=0.750 m/s[/tex]

[tex]s=6m[/tex]

First determinate the time using equation of Newton's law

[tex]d=\frac{uf+ui}{2}*t[/tex]

uf=0 solve to t'

[tex]t=\frac{2*d}{u_i}=\frac{2*6m}{0.75m/s}[/tex]

[tex]t=16s[/tex]

The impulse concept tells about the force in terms of velocity so:

[tex]F=m*a[/tex]

[tex]a=\frac{u'}{t'}[/tex]

[tex]F_i=m*\frac{u}{t}=1x10^7kg\frac{0.750m/s}{16s}[/tex]

[tex]F_i=4.6875x10^5 N[/tex]

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