A tugboat pulls a ship with a constant net horizontal force of 5.00 × 103 N and causes the ship to move through a harbor. How much work is done on the ship if it moves a distance of 3.00 km?

The work done on the ship when a tugboat pulls it with a constant net horizontal force of 5.00x10³ N over a distance of 3.00 km is 15.0x10⁶ J.  

We can find the work done on the ship as follows:

(1)

Where:

F: is the net horizontal force = 5.00x10³ N

d: is the displacement = 3.00 km = 3000 m

By entering the above values into equation (1), we have:

Therefore, the work done on the ship is 15.0x10⁶ J.

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15 MW

Explanation:

Work done is a product of force and distance moved.

W=fd where f represnt force while d is distance

Substituting f with 5 kN and 3 000 for d then

W= 5*3000=15000 kW which is equivalent to 15 MW

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force