A tugboat pulls a ship with a constant horizontal force of 6000N and causes the ship to move through a harbor. The water exerts a constant drag force on the ship of 500N. How much work is done on the ship if it moves a distance of 4.0 km?
Force is causing parallel displacement. Work is done by boat. Work by boat is acting positively. The boat has kinetic energy by moving forward on it's own. Work has no potential. Is work by kinetic energy is (0.5)mv^2? But I don't know the mass. I know the exerted force is 6000. Force is mass and acceleration. If force is the dividend of acceleration, I could find mass of boat. But what is acceleration? change in velocity partitioned by the time of change. But what is velocity? It is not given. Go back. Perhaps the belief that work is done by KE is erroneous or irrelevant? (Which is it?) Without the knowledge of speed, rate of change, and mass, how can we find out what work is? What is work? Well it's measured in joules... What is a joule? Ah, something like force partitioned by time. What is force? Some newtons. What is time? Some seconds. (SI units). So to know Work, is to know partitioned newtons of force. I know That the boat is exerting 6000 newtons in a (assumed) parallel direction of displacement. Parallel as in, 100% of the force is delivered in displacement. We have 100% of displacement. What is displacement? 4.0 km. 100% of 4.0km is 4.0km. So 6000 Newtons of force is delivered 4.0 (km). For 4.0 (km), 6000 newtons of force has been delivered. 4.0 (km) and 6000 (N) form a product of [24,000 (km)(N)]
A newton is the product of 1 kg with the quotient of the displacement (SI meter) partitioned by a squared second. A newton is the product of 1 kg and an acceleration; the product of mass and acceleration.
The product, 24,000, should have units kg, m, and s. This product does not have these units. The multiplicand, displacement of the boat was measured in km. 4.0 km is simply 4 thousand meters. To correct my mistake, 4,000 (meters) and 6,000 (newtons) form the product 2.4(10^7) [1kg(m^2)/(s^2)]
Because the nature of this product's units is such that mass, displacement, and time intertwine, these units make up the joule. The product is 2.4(10^7) Joules. 2.4(10^7) Joules of work by Kinetic Energy so I have not made an erroneous error I believe.
I have found that, in this case, W= F * d * amount of force delivered, determined by the angle at which the do-er does work to displace.
So, because I have to take into account the hindrance of water to the boat's positive work, do I have to supply this hindrance as a negative energy (in joules) to the positive work being done? In other words, Do I deduct the water's exertion from that of the boat's? or do I add these two energies together and treat on or the other as a positive or negative value?
work done by tugboat = 6000 * 4000
= 24 * 10^6 Joules
work done by water = -500*4000 = -2*10^6 Joules (negative because force is opposite to motion)
net work done on ship = 22*10^6 Joules
Note unbalanced force on ship leads to acceleration but we do not care.
a Joule is a Newton meter
which is
kg meter/s^2 * meter = kg meter^2/s^2
To calculate the work done on the ship, you need to consider both the force applied by the tugboat and the drag force exerted by the water. The work done by the tugboat is positive, while the work done by the water drag is negative, as it opposes the motion of the ship.
The formula for work is given by W = F * d * cos(theta), where F is the force applied, d is the distance moved, and theta is the angle between the force and displacement vectors.
In this case, the force applied by the tugboat is 6000N and the distance moved is 4.0 km. However, we need to convert the distance to meters to match the units of force. So, 4.0 km is equal to 4000 m.
The angle between the force and displacement vectors is not given, but since it is stated that the force is causing parallel displacement, we can assume that theta is 0 degrees. In this case, cos(0) = 1, so the cosine term is equal to 1.
Therefore, the work done by the tugboat is W = 6000N * 4000m * 1 = 24,000,000 Joules.
Now, let's consider the drag force exerted by the water. The drag force is given as 500N, and since it opposes the motion, the angle between this force and the displacement is 180 degrees. The cosine term becomes cos(180) = -1.
So, the work done by the water drag is W = 500N * 4000m * -1 = -2,000,000 Joules.
To find the total work done on the ship, you need to add the work done by the tugboat and the work done by the water drag: 24,000,000 Joules + (-2,000,000 Joules) = 22,000,000 Joules.
Therefore, the total work done on the ship as it moves a distance of 4.0 km is 22,000,000 Joules.
Note: The work done by kinetic energy equation (W = 0.5 * m * v^2) is not applicable in this case since you do not have information about the mass or velocity of the ship. We followed the approach of calculating work using the applied force and displacement.