# application to physics and engineering

• Nov 17th 2007, 05:40 PM
chocolatelover
application to physics and engineering
Hi everyone,

A spring has a natural length of 20 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

I know that work=(force)(distance)

Force=25 N
distance=25-20=5
work=(25N)(5cm)

Is this correct so far? If it isn't, could you please show me what to do?

Thank you very much
• Nov 17th 2007, 08:47 PM
earboth
Quote:

Originally Posted by chocolatelover
A spring has a natural length of 20 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

I know that work=(force)(distance)

Force=25 N
distance=25-20=5
work=(25N)(5cm)

Is this correct so far? ...

Hello,

unfortunately no.

1. If you stretch a spring the necessary force is not constant but is proportional to the distance you have stretched the spring:

If F is the force, d is the distance of stretching then:

$\displaystyle \frac Fd=k$ . This quotient is constant (in certain boarders: You can't stretch a spring of 10 cm to a length of 1 km). In your case:

$\displaystyle \frac{25\ N}{10\ cm}=\boxed{k=2.5\ \frac N{cm}}$

2. Since the force is not constant your formula to calculate the work is not valid. The work done to stretch a spring can be calculated by:

$\displaystyle w=\frac12 \cdot k \cdot d^2$ . Plug in the values for k and d.

$\displaystyle w=\frac12 \cdot 2.5\ \frac N{cm} \cdot 5^2\ cm^2=31.25\ Ncm = 0.3125 \ J$
• Nov 18th 2007, 10:43 AM
chocolatelover
Thank you very much

I haven't learned that formula

Could you also do it this way?

20cm to 30 cm
change=
.10m

f(.10)=25
.10k=25
k=250

f(x)=kx=250dx
w=int. .10 to .25(250xdx)
=(250x^2/2)from .10 to .25 (Are these limits correct?)
w=6.5625J

Thank you very much
• Nov 18th 2007, 01:50 PM
topsquark
Quote:

Originally Posted by chocolatelover
Thank you very much

I haven't learned that formula

Could you also do it this way?

20cm to 30 cm
change=
.10m

f(.10)=25
.10k=25
k=250

f(x)=kx=250dx
w=int. .10 to .25(250xdx)
=(250x^2/2)from .10 to .25 (Are these limits correct?)
w=6.5625J

Thank you very much

Actually, the two of you are using the same method:
$\displaystyle W = \int \vec{F} \cdot \vec{ds}$

When you put Hooke's law in for the force, you get the formula earboth used.

-Dan
• Nov 18th 2007, 03:40 PM
chocolatelover
Do you see where I made my mistake?

Thank you very much