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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.

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Presentation on theme: "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."— Presentation transcript:

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 SoT = =  = force required to double the length of string

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

3 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. e = 1.2 T 2g A In equilibrium So length string = 4.2m in equilibrium

4 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 Solvinga = 2.9ms –2

5 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 = 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 = 52.5 =  = 1350

7 Energy stored in a stretched elastic string If a constant force moves a distance x then the work done = F  s W = F  s i.e. area under the line Stretching a string needs an increasing force. F s 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 The ball is now lifted up to the point A and dropped. T 2g A Find the velocity at the equilibrium point and the max length of string. 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. e = 1.2 In equilibrium So length string = 4.2m in equilibrium

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

11 N = natural length = 3m A E = equilibrium length = 4.2m 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

12 N = natural length = 3m A 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. 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 B

13 N A B At APE = 2g  4.2 KE = 0 EPE = 0 Make the equilibrium point have zero P.E. So points below this point have negative P.E. E

14 N A B At APE = 2g  4.2 KE = 0 EPE = 0 Make the equilibrium point have zero P.E. So points below this point have negative P.E. PE = 0 KE =   2  v 2 EPE = PE = -2gh KE = 0 EPE = h1.2m E

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

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

17 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. e = 0.4m T 2g A In equilibrium So length string = 1.4m in equilibrium

18 N A At APE = 2g  1.4 KE = 0 EPE = 0 E PE = 0 KE =   2  v 2 EPE = PE = -2g  0.2 KE = 0 EPE = initial acc.  T – 2g = 2a subst x = 0.6 and  = 50 a =  (50  ) = 5ms -2 B

19 N A At APE = 2g  1.4 KE = 0 EPE = 0 E PE = 0 KE =   2  v 2 EPE = PE = -2g(0.6) KE = 0 EPE = 0.2 Energy at B = Energy at E 1 = v 2 v = 1ms So velocity at E = 1ms -1 B -2g(0.6) + = 0 +   2  v 2 +

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

21 N A B At APE = 2g  1.4 KE = 0 EPE = 0 PE = 2g  (0.4-x) KE = 0 EPE = PE = -2g(0.2) KE = 0 EPE = 0.2 Energy at B = Energy at M x = 0.2m or 0.6m But 0.6m is at the bottom so x = 0.2m x M at M -2g(0.2) + = 2g  (0.4- x) + Finding the maximum height – M E So the length of the spring at the max height is 1.2m i.e 0.2m above E

22 N A B At APE = 2g  1.4 KE = 0 EPE = Energy at B = Energy at M x = 0.2m or 0.6m But 0.6m is at the bottom so x = 0.2m x M -2g(0.2) + = 2g  (0.4-x) + Finding the maximum height – M So the length of the spring at the max height is 1.2m i.e 0.2m above E


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