Objective: You need to drag this whale 10 m back into the sea….. a) Use cons of

Question

Objective: You need to drag this whale 10 m back into the sea…..

a) Use cons of momentum to determine the speed of the system right after the rope is taut.

b) Use impulse-momentum to determine the friction force.

c) Then find the coefficient of friction between the whale and the sand.

d) Then use system energy to determine how far he can drag the whale with one try.

e) Then determine how many tries it takes to get the whale back into the ocean.

·15-17· The 55-Mg humpback whale is stuck on the shore due to changes in the tide. In an effort to rescue the whale, a 12-Mg tugboat is used to pull it free using an inextensible rope tied to its tail. To overcome the frictional force of the sand on the whale, the tug backs up so that the rope becomes slack and then the tug proceeds forward at 3 m/s. If the tug then turns the engines off, determine the average frictional force F on the whale if sliding occurs for 1.5 s before the tug stops after the rope becomes taut.

Explanation / Answer

Solution:

a) Momentum of whale initially = p1 =0 since it is at rest

MOmentum of boat initially = p2 = M2v2 = ( 12 x10^6 * 10^-3) (3) = 36000 kg m/s

FInal momentum = (m1+m2)*V

from the momentum conservation , p2 = (m1+m2) V

=> V = 36000 /(12000+ 5500) = 2.1 m/s = final velocity of the whale + boat

b) t= time = 1.5 s

initial momentum of whale = 0

final momentum = 36000 kg m/s

force x time = change in momentum

force = 36000 / 1.5 = 24000 N

FrictionForce = 24000 N 24 kN

c) Coefficient of friction = 24000 / (12000+5500) * 9.8 = 0.14

the boat's mass is also contributing to the normal force since they move together to overcome the friction .

d) Acceleration = ma = umg

=> a = ug = 0.14 * 9.8 = 1.37 m/s^2

e) the displacement of the whale boat system = d = v^2/2a = (2.1)^2 / 2*1.37 = 1.61 m

No. of tries = N = 10 m /1.61m = 6 tries

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