Question

A large cruise ship of mass 7.00 ✕ 107 kg has a speed of 11.6 m/s at some instant. (a) What is the ship's kinetic energy at this time? (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.) (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 3.00 km?

Answer 1

(a)
`4.7\cdot 10^9 J`

The kinetic energy of an object is given by

`K=(1)/(2)mv^2`

where

m is the mass of the object

v is its speed

For the ship in this problem,

`m=7.00 \cdot 10^7 kg` is the mass

`v = 11.6 m/s` is the speed

Substituting,

`K=(1)/(2)(7.00 \cdot 10^7 kg)(11.6 m/s)^2=4.7\cdot 10^9 J`

(b)
`-4.7\cdot 10^9 J`

The work done on the ship is equal to the change in kinetic energy of the ship; so we have:

`W = K_f - K_i`

where

W is the work done

Kf is the final kinetic energy of the ship, which is zero since it comes to rest

Ki is the initial kinetic energy of the ship, which is
`4.7\cdot 10^9 J`

Substituting into the formula,

`W=0-4.7\cdot 10^9 J=-4.7\cdot 10^9 J`

and the sign is negative because the force used to stop the ship acts against the direction of motion of the ship, in order to slow it down.

(c)
`1.57\cdot 10^6 N`

The work done by the force applied to stop the ship is

W = Fd

where

W=-4.7\cdot 10^9 J is the work done

F is the force applied

d = 3.00 km = 3000 m is the displacement of the ship

Solving the equation for F, we find the force:

`F=(W)/(d)=(-4.7\cdot 10^9 J)/(3000 m)=-1.57\cdot 10^6 N`

and ignoring the negative sign, the magnitude of the force is

`1.57\cdot 10^6 N`