Question

A large cruise ship of mass 6.00 ✕ 10⁷ kg has a speed of 12.8 m/s at some instant.

(a) What is the ship's kinetic energy at this time?
J

(b) How much work is required to stop it? (Give the work done on the ship. Include the sign of the value in your answer.)
J

(c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 3.90 km?
N

Answer 1

given data

mass m= 6.0 ✕ 10^7 kg and speed is v = 12.8 m/s

a)

we know that kinetic energy is given by = 1/2 m x v^2

KE = 0.5x6.0x10^7 kg x12.8 m/s x 12.8 m/s = 4.915x10^9  joules

KE = 4.915x10^9  joules

b )

the work done is equal to the change in K.E. of the object! This is the Work-Energy theorem or the relation between Kinetic energy and Work done. In other words, the work done on an object is the change in its kinetic energy. W = Δ(K.E.) so work done is negative sign of kinetic energy so

W= - 4.915x10^9  joules

c)

The work is calculated by multiplying the force by the amount of movement of an object

W = F ✕ S.......... SO FORCE IS F = W/S

given data

S = 3.90 km =3.9✕10^3 m AND   w= - 4.915x10^9  joules

F = - 4.915x10^9  J / (3.9✕ 10^3 m) = -1.26 x10^6 N

magnitude of force is =1.26 x10^6 N