A supertanker (mass = 1.57E+8 kg) is moving with a constant velocity. Its engines generate a forward thrust of 7.47E+5 N. Determine the magnitude of the resistive force exerted on the tanker by the water.
Oh come on, the velocity is constant therefore net force = 0
When we typed in 0, it said it was the incorrect answer
Good grief ! The NET force is zero.
Yes, the net force is 0, but the book did not ask for the net force; it asked for the force exerted on the tanker by the water.
M*g = 1.57*10^8 * 9.8 = 15.4*10^8N. = Wt. of the tanker(Ft) = Normal force(Fn).
Fr = u*Fn = 15.4*10^8u. = Resistive force of the water.
Ft + Fr = M*a.
Ft + Fr = M*0
Ft + Fr = 0.
Fr = -Ft = -15.4*10^8 N.
Correction: Fr = -Ft = -7.47*10^5N.
To determine the magnitude of the resistive force exerted on the tanker by the water, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration.
In this case, since the tanker is moving with a constant velocity, we know that the acceleration is zero. Therefore, the net force acting on the tanker is also zero, as there is no change in velocity.
The net force acting on the tanker is the sum of the forward thrust generated by the engines and the resistive force exerted by the water:
Net force = Thrust - Resistive force
Since the net force is zero and the thrust is given as 7.47E+5 N, we can rearrange the equation to solve for the resistive force:
Resistive force = Thrust
Therefore, the magnitude of the resistive force exerted on the tanker by the water is 7.47E+5 N.