1. Elasticity
Hooke's Law : the extension in an elastic string is proportional to the applied force .
T = 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 = = 
 = force required to double the length of string

2.  mg Elastic Strings / Springs In equilibrium  T = mg T =
The equilibrium extension is often called e
 mg

3. 2g Ex1 A 2kg ball is attached to a 3m string of l=50N.
The string is attached to a point A.
The ball hangs in equilibrium. Find the length of the string. T 2g A
In equilibrium
e = 1.2
So length string = 4.2m in equilibrium

4. A mass of 2kg is hung from a 50cm string with  = 30N
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.
In equilibrium T – 2g = 0 so T = = 2g
Solve for e e = = 0.33m
e = equilibrium extension 2g T
When it is pulled down 10 cm it is no longer in equilibrium so use resultant force = ma
T – 2g = 2a but T = So – 2g = 2a
Solving a = 2.9ms–2

5. Find the force exerted by the jaws of the vice.
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 = T =

6. 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 =
Using T = =   = 1350

7. Energy stored in a stretched elastic stringF
If a constant force moves a distance x then the work done = F  s
W = F  s
i.e. area under the line s
Stretching a string needs an increasing force. F s
So the work done stretching a string a distance x is : W =

8. 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
E.P.E =
So now the total energy equation is :
Total mechanical Energy = K.E + P.E + E.P.E

9. 2g A 2kg ball is attached to a 3m string of l=50N.
The string is attached to a point A.
The ball hangs in equilibrium. Find the length of the string.
In equilibrium
e = 1.2
So length string = 4.2m in equilibrium T 2g A
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.

10. A
From A to N the body accelerates as there is no upward force as it falls.
N = natural length = 3m

11. 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

12. 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
From E to B the body decelerates as the upward force is now greater than the downward force. It stops at B. B

13. A
At A PE = 2g KE = EPE = 0
Make the equilibrium point have zero P.E.
So points below this point have negative P.E. N E B

14. A
At A PE = 2g KE = EPE = 0
Make the equilibrium point have zero P.E.
So points below this point have negative P.E. N
1.2m E
PE = KE = 2v EPE = h
PE = -2gh KE = 0 EPE = B

15. A
At A PE = 2g KE = 0 EPE = 0
Energy at A = Energy at E
2g4.2 = 0 + 2v Solve for v g = 10ms-2
0 = v2 – 72
v = 8.5ms-1 N E
PE = KE = 2v2 EPE = h B
PE = -2gh KE = EPE =

16. A
At A PE = 2g KE = 0 EPE = 0
Energy at A = Energy at B
2g4.2 = -2gh +
0 = h2 – 72
Solve quadratic for h
h = 2.93m So total distance = 7.13m (add on 4.2) N E
PE = KE = 2v2 EPE = h
PE = -2gh KE = EPE = B

17. 2g Ex 2 Mass 2kg hung from a 1m spring with l = 50N.
Find the extension. It is then pulled down a further 20cm.
Find initial acceleration, velocity at equilibrium point and max height. T 2g A
In equilibrium
e = 0.4m
So length string = 1.4m in equilibrium

18. A
At A PE = 2g KE = 0 EPE = 0
initial acc.  T – 2g = 2a
subst x = 0.6 and l = 50
a =  (50 ) = 5ms-2 N
0.4 E
PE = KE = 2v EPE =
0.2
PE = -2g  KE = EPE = B

19. A
At A PE = 2g KE = 0 EPE = 0
Energy at B = Energy at E
-2g(0.6) = 0 + 2v2 +
1 = v2
v = 1ms-1
So velocity at E = 1ms-1 N
0.4 E
PE = 0 KE = 2v2 EPE =
0.2
PE = -2g(0.6) KE = EPE = B

20. A
At A PE = 2g KE = 0 EPE = 0
Energy at B = Energy at N
-2g(0.2) = 2g0.4 + 2v2
v =-3 impossible so it does not reach N
PE = 2g KE = 2v2 EPE = 0
N = 1m
0.4
E = 1.4m
PE = KE = 2v2 EPE =
0.2
PE = -2g KE = EPE =
B = 1.6

21. Finding the maximum height – M
At A PE = 2g KE = 0 EPE = 0
Energy at B = Energy at M
-2g(0.2) = 2g(0.4- x) +
x = 0.2m or 0.6m
But 0.6m is at the bottom so x = 0.2m N x
PE = 2g(0.4-x) KE = EPE =
M at M E
0.2
PE = -2g(0.2) KE = EPE = B
So the length of the spring at the max height is 1.2m i.e 0.2m above E

22. Finding the maximum height – M
At A PE = 2g KE = 0 EPE = 0
Energy at B = Energy at M
-2g(0.2) = 2g(0.4-x) +
x = 0.2m or 0.6m
But 0.6m is at the bottom so x = 0.2m N x
So the length of the spring at the max height is 1.2m i.e 0.2m above E M
0.2 B