A $120 k g$ dolphin decelerates from $15 m / s$ to $7.5 m / s$ in $5 s$ to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (Assume that gravitational force is balanced by the buoyant force of the water)

Option 1 - <math> <mn>120</mn> <mi> </mi> <mi>N</mi> </math>

Option 2 - <math> <mn>18</mn> <mi> </mi> <mi>N</mi> </math>

Option 3 - <math> <mn>180</mn> <mi> </mi> <mi>N</mi> </math>

Option 4 - <math> <mn>60</mn> <mi> </mi> <mi>N</mi> </math>

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Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 3

Detailed Solution:

From equation of uniformly accelerated motion

$v = u + a t$

$7.5 = 15 + 5 a$

$a = - 1.5 m / s^{2}$

$n o w F = m a$

$|\overset{?}{F}| = 120 * 1.5$

$= 180 N$

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A 120 k g dolphin decelerates from 15 m / s to 7.5 m / s in 5 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (Assume that gravitational force is balanced by the buoyant force of the water)

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