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.
5 answers
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.
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.
1 answer