#### Transcript Elastic Strings - A Level Maths Help

```Elasticity
Hooke's Law : the extension in an elastic string is proportional to
the applied force .
T=
l x
l
x = extension l = natural length
= modulus of elasticity and depends on the material involved
NOT the length.
If the string is doubled in length then the extension will equal the
original length l
So
T=
ll
l
=
 = force required to double the length of string
Elastic Strings / Springs
In equilibrium  T = mg
l x
= mg
l
x =
m gl
l
T=
l x
 mg
l
The equilibrium extension is often called e
e =
m gl
l
Ex1 A 2kg ball is attached to a 3m string of =50N.
The string is attached to a point A.
The ball hangs in equilibrium. Find the length of the string.
A
T
In equilibrium
50e
T = l e=
= 2g
l
3
e = 1.2
So length string = 4.2m in equilibrium
2g
A mass of 2kg is hung from a 50cm string with  = 30N. It is
pulled down 10cm from the equilibrium position find its initial
acceleration.
T
In equilibrium T – 2g = 0
Solve for e
e=
T = l e = 2g
so
2g ´ 0.5
30
l
= 0.33m
e = equilibrium extension
2g
When it is pulled down 10 cm it is no longer in equilibrium so use
resultant force = ma
T – 2g = 2a
but T = l x
30 ´ 0.43
- 2g = 2a
0.5
l
Solving
So
a = 2.9ms–2
l x
– 2g = 2a
l
T=
Springs
The formula also applies if a spring is either stretched or
compressed.
Ex.1 An elastic spring of modulus of elasticity 30N and natural
length 20cm is compressed in a vice to a length of 17cm.
Find the force exerted by the jaws of the vice.
The compression in the spring is the same throughout and is
equal to the force exerted by the jaws of the vice
Using T =
l x
l
T = 30 ´
0.03
0.2
= 4.5N
Ex.2
An 18cm deep mattress containing 14 springs is compressed by
0.7cm when a 75kg sleeper lies on it.
Find the force to compress it 1cm
If a force of 75g is applied to the mattress then each spring
experiences a compression force of:
F =
75g
= 52.5N
14
Using T =
l x
l
52.5 =
l ´ 0.007
0.18
  = 1350
Energy stored in a stretched elastic string
If a constant force moves a distance x
then the work done = F  s
F
s
W=Fs
i.e. area under the line
Stretching a string needs an
increasing force.
F
W or k = a r ea u n d er lin e
s
=
ò F ds = ò T ds = ò
λs
λs
2
ds =
l
So the work done stretching a string a distance x is : W = l x
2l
2l
2
But in stretching a string then energy is acquired so:
The Elastic energy acquired = the work done stretching the string
The Elastic energy is called Elastic Potential energy E.P.E
2
l
x
E.P.E =
2l
So now the total energy equation is :
Total mechanical Energy = K.E + P.E + E.P.E
A 2kg ball is attached to a 3m string of =50N.
The string is attached to a point A.
The ball hangs in equilibrium. Find the length of the string.
In equilibrium
50e
T = l e=
= 2g
l
3
e = 1.2
A
T
2g
So length string = 4.2m in equilibrium
The ball is now lifted up to the point A and dropped.
Find the velocity at the equilibrium point and the
max length of string.
A
From A to N the body accelerates as there is no
upward force as it falls.
N = natural length = 3m
A
From A to N the body accelerates as there is no
upward force as it falls.
From N to E the body continues to accelerate as the
downward force is greater than the upward force.
The body has max velocity at E as at this point the
upward and downward forces balance
N = natural length = 3m
E = equilibrium length = 4.2m
A
From A to N the body accelerates as there is no
upward force as it falls.
From N to E the body continues to accelerate as the
downward force is greater than the upward force.
The body has max velocity at E as at this point the
upward and downward forces balance
N = natural length = 3m
E = equilibrium length = 4.2m
B
From E to B the body decelerates as the upward
force is now greater than the downward force. It
stops at B.
A At A
PE = 2g4.2
KE = 0
EPE = 0
Make the equilibrium point have zero P.E.
So points below this point have negative P.E.
N
E
B
A
At A
PE = 2g4.2
KE = 0
EPE = 0
Make the equilibrium point have zero P.E.
So points below this point have negative P.E.
N
1.2m
E
PE = 0
KE =
2v2
EPE
h
B
PE = -2gh
KE = 0
EPE =
50 ´ 1.2 2
= 2´ 3
50( h + 1.2) 2
2´ 3
A
At A
PE = 2g4.2
KE = 0 EPE = 0
Energy at A = Energy at E
2g4.2 = 0 + 2v2 +
50 ´ 1.2 2
2´ 3
Solve for v g = 10ms-2
0 = v2 – 72
v = 8.5ms-1
N
E
PE = 0
KE =
2v2
EPE =
h
B
PE = -2gh
KE = 0
EPE =
50 ´ 1.2 2
2´ 3
50( h + 1.2) 2
2´ 3
A
At A
PE = 2g4.2
KE = 0 EPE = 0
Energy at A = Energy at B
2g4.2 = -2gh
50( h + 1.2) 2
+ 2´ 3
0 = 50 h2 – 72
6
h = 2.93m
So total distance = 7.13m (add on 4.2)
N
h
E
B
PE = 0
PE = -2gh
KE =
2v2
KE = 0
EPE =
EPE
50 ´ 1.2 2
2´ 3
50( h + 1.2) 2
= 2´ 3
Ex 2 Mass 2kg hung from a 1m spring with  = 50N.
Find the extension. It is then pulled down a further 20cm.
Find initial acceleration, velocity at equilibrium point and max height.
A
T
In equilibrium
50e
T = l e=
= 2g
l
1
e = 0.4m
So length string = 1.4m in equilibrium
2g
A
At A
PE = 2g1.4
KE = 0 EPE = 0
initial acc.  T – 2g = 2a
subst x = 0.6 and = 50
50 ´ 0.6 - 2g = 2a
1
a =  (500.6 - 20) = 5ms-2
0.4
N
E
PE = 0
KE =
2v2
EPE =
50 ´ 0.4 2
2´ 1
EPE =
5 0( 0 .6 ) 2
2´ 1
0.2
B
PE = -2g 0.2 KE = 0
A
At A
PE = 2g1.4
KE = 0 EPE = 0
Energy at B = Energy at E
-2g(0.6) +
1 = v2
v = 1ms-1
0.4
50 ´ 0.6 2
2´ 1
= 0 + 2v2 +
50 ´ 0.4 2
2´ 1
So velocity at E = 1ms-1
N
E
PE = 0 KE = 2v2 EPE =
0.2
B
PE = -2g(0.6) KE = 0
50 ´ 0.4 2
2´ 1
EPE =
5 0( 0 .6 ) 2
2´ 1
A
At A
PE = 2g1.4
KE = 0 EPE = 0
Energy at B = Energy at N
-2g(0.2) +
50 ´ 0.6 2
2´ 1
= 2g0.4 + 2v2
v =-3 impossible so it does not reach N
0.4
N = 1m
E = 1.4m
PE = 2g0.4 KE = 2v2 EPE = 0
PE = 0
KE =
2v2
EPE
50 ´ 0.4 2
= 2´ 1
0.2
B = 1.6
PE = -2g0.2 KE = 0
EPE =
5 0( 0 .6 ) 2
2´ 1
Finding the maximum height – M
A
At A
PE = 2g1.4
KE = 0 EPE = 0
Energy at B = Energy at M
-2g(0.2) +
50 ´ 0.6 2
2´ 1
= 2g(0.4- x) +
50 ´ x 2
2´ 1
x = 0.2m or 0.6m
But 0.6m is at the bottom so x = 0.2m
x
0.2
N
M
E
B
at M PE = 2g(0.4-x) KE = 0 EPE =
PE = -2g(0.2) KE = 0
EPE =
50 ´ x 2
2´ 1
5 0( 0 .6 ) 2
2´ 1
So the length of the spring at the max height is 1.2m i.e 0.2m above E
Finding the maximum height – M
A
At A
PE = 2g1.4
KE = 0 EPE = 0
Energy at B = Energy at M
-2g(0.2) +
50 ´ 0.6 2
2´ 1
= 2g(0.4-x) + 502 ´´
x2
1
x = 0.2m or 0.6m
But 0.6m is at the bottom so x = 0.2m
x
N
M
0.2
B
So the length of the spring at the max
height is 1.2m i.e 0.2m above E
```