Question

Two tugboats pull a disabled container-ship loaded to 40,000 metric tons (40 million kg). Each tug exerts a constant force of 1.80×10⁶ N , one at an angle 12.0 ∘ west of north, and the other at an angle 12.0 ∘ east of north, as they pull the ship 0.880 km northward at constant 6 m/s.

A. What is the total work done to the ship by the two tugboat Forces? in mega joules

B. What is the container-ship's Kinetic Energy? in mega joules

C.Starting from rest, how far would the tugs have pulled the ship before achieving its 6 m/s velocity? Ignore water resistance for this estimate.

Answer 1

Net force in north direction = 2 X 1.80 X 10^6N X cos12 = 3.52 Mega N

Distance tugged = 0.880 X 10^3 meter North

Work done by tugs = 3.52 X 10^6 N X 0.880 X 10^3m = 3097 X 10^6 J = 3097 Mega J

B. KE = 0.5 X 40 X 10^6 Kg X (6m/s)^2 = 720 Mega J

C. Force on ship = 3.52 X 10^6 N ( towards north)

Acceleration in north direction = 3.52 X 10^6 N/(40X10^6 Kg) = 0.088 m/s^2

Final velocity = 6 m/s Initial velocity = 0m/s

6^2 = 2 X 0.88 m/s^2 X S , where S is the distance moved

S = 20.4 m