Question

For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 7.45 105 N. What is the magnitude of the resistive force exerted by the water on the cruise ship

Answer 1

Answer:

7.45 10^5N.

Explanation:

according to newtons second law of motion;

$\sum F_x = ma_x\\F_{app} - F_r = ma_x$

Fapp is the applied force

Fr is the resistive force

m is the mass of the luxury

a is the acceleration

Since the huge luxury liner move with constant velocity, then acceleration is zero i.e a = 0. The equation becomes;

$F_{app} - F_f = m(0)\\F_{app} - F_f =0\\F_{app} = F_f$

This shows that the applied force will be equal to the resistive force if the velocity is constant.

Given Fr = 7.45 10^5 N therefore the resistive force will also be 7.45 10^5N.

Hence the magnitude of the resistive force exerted by the water on the cruise ship is 7.45 10^5N.