Thread: Science: work and force

1. X lifts a 500g mug of beer to a height of 0.5 m. How much work does X do?

2. Y lifts a 1 kg mug to a height of 2.5m in 5 seconds. How much work does Y do?

3. Ship = 4 500 tonnes, engine provides thrust force of 90 000N. What is teh shortest time in which teh ship can accelerate from 10-28 kms/hr?


I don't really get the method and formula we have to use to get this. Thanks to whoever works this out in advance =D.

Originally Posted by delicate_tears

1. X lifts a 500g mug of beer to a height of 0.5 m. How much work does X do?

The work done is equal to the change in potential energy which is the
mass times the chage in height times the acceleration due to gravity.

This comes to the same thing as the force times the distance it moves
through as the force is the mass times the acceleration due to gravity
and the distance moved through is the change in height.

2. Y lifts a 1 kg mug to a height of 2.5m in 5 seconds. How much work does Y do?

The time here is a noise parameter it has nothing to do with the work done
which is calculated in exactly the same way as in part (1).

3. Ship = 4 500 tonnes, engine provides thrust force of 90 000N. What is teh shortest time in which teh ship can accelerate from 10-28 kms/hr?

To do this we ignore drag terms and so:

force=mass x acceleration.

Here we have a constant mass 4500 tonnes=4500000 kg, the acceleration
will also be constant, and so acceleration = change in velocity/time
(you will have to convert the velocities into m/s).

The force is the thrust 90000N.

Plug these into the first equation and solve for time.

RonL