# Physical work done by force

• Jan 21st 2007, 05:51 AM
totalnewbie
Physical work done by force
Weight of body 1 Kg
Within 5 seconds the force that influenced the body was 20 Newtons.

How much is physical work done by force ?
• Jan 21st 2007, 08:26 AM
CaptainBlack
Quote:

Originally Posted by totalnewbie
Weight of body 1 Kg
Within 5 seconds the force that influenced the body was 20 Newtons.

How much is physical work done by force ?

There is not enough information here to determine the work done without
making some assumptions about what has been omitted from the question.

1. Assume you mean the body has mass of 1kg

2. Assume the only force acting on the body is one of 5 N, acting for 5s

3. Assume the body starts at rest.

Then we have constant accelleration:

$\displaystyle a=f/m$,

so the velocity is:

$\displaystyle v=(f/m)t$

displacement from initial position is:

$\displaystyle s=(f/[2m])t^2$

Therefore with $\displaystyle f=20\ \mbox{N}$, $\displaystyle m=1 \mbox{ kg}$, and $\displaystyle t=5 \mbox{ s}$, we have:

$\displaystyle s=10 \times 25=250\mbox{ m}$

Therefore the work done is the force times the displacement $\displaystyle =5000 \mbox{\ Joules}$

RonL
• Jan 21st 2007, 08:54 AM
totalnewbie
Thanks!

But I started to think why gravitational forces are not involved ?
For exmaple: $\displaystyle F=ma+-mg$
Despite of constant acceleration gravitational forces $\displaystyle F=mg$ influence that body.
• Jan 21st 2007, 10:45 AM
CaptainBlack
Quote:

Originally Posted by totalnewbie
Thanks!

But I started to think why gravitational forces are not involved ?
For exmaple: $\displaystyle F=ma+-mg$
Despite of constant acceleration gravitational forces $\displaystyle F=mg$ influence that body.

Well this could be taking place where the acceleration due to gravity has
negligible effect over the time intervals in question.

Alternatively it could be conducted on a low friction surface, where the
normal reaction from the surface balances the gravitational force.

RonL
• Jan 22nd 2007, 02:20 PM
topsquark
Quote:

Originally Posted by CaptainBlack
Therefore the work done is the force times the displacement $\displaystyle =5000 \mbox{\ watts}$

Owie! The unit for work is "Joules." :)

-Dan
• Jan 22nd 2007, 07:35 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Owie! The unit for work is "Joules." :)

-Dan

Oppss..