elasticity hooke's law : the extension in an elastic string is proportional to the applied...

ElasticityHooke'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

xll

lll

Elastic Strings / Springs

In equilibrium T = mg

xmg

l=

l

mg

T = xll

mgx =

ll

The equilibrium extension is often called emg

e=l

l

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.

e = 1.2

T

2g

A

50 eT 23e gl= = =lIn equilibrium

So length string = 4.2m in equilibrium

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

ell

2g 0.530´

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

xll x

ll

30 0.43 2 20.5

g a´ - =

SpringsThe 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 = xll 30 0.03

4.5N0.2´

=

T =

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

52.5N14

=

xll 0.007

0.18l ´

Using T = 52.5 = = 1350

Energy stored in a stretched elastic string

λ λ

l l

2s s

Fds Tds ds2

Work = area under line

= = =ò ò ò=

If a constant force moves a distance x then the work done = F s

W = F si.e. area under the line

Stretching a string needs an increasing force.

F

s

F

s2x

2l

lSo the work done stretching a string a distance x is : W =

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 = 2x

2l

l

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

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 l=50N. The string is attached to a point A. The ball hangs in equilibrium. Find the length of the string.

e = 1.2

50 eT 23e gl= = =lIn equilibrium

So length string = 4.2m in equilibrium

N = natural length = 3m

AFrom A to N the body accelerates as there is no upward force as it falls.

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

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

N

A

B

At A PE = 2g4.2 KE = 0 EPE = 0

Make the equilibrium point have zero P.E. So points below this point have negative P.E.

E

N

A

B

At A PE = 2g4.2 KE = 0 EPE = 0

Make the equilibrium point have zero P.E. So points below this point have negative P.E.

PE = 0 KE = 2v2 EPE = 250 1.2

2 3´´

PE = -2gh KE = 0 EPE = 250 h+1.2)(

2 3´

h

1.2mE

N

A

B

At A PE = 2g4.2 KE = 0 EPE = 0

E PE = 0 KE = 2v2 EPE = 250 1.2

2 3´´

PE = -2gh KE = 0 EPE = 250 h+1.2)(

2 3´

h

Energy at A = Energy at E

2g4.2 = 0 + 2v2 + Solve for v g = 10ms-2250 1.22 3´´

0 = v2 – 72v = 8.5ms-1

N

A At A PE = 2g4.2 KE = 0 EPE = 0

E PE = 0 KE = 2v2 EPE = 250 1.2

2 3´´

PE = -2gh KE = 0 EPE = 250 h+1.2)(

2 3´

h

Energy at A = Energy at B

2g4.2 = -2gh +250 h +1.2)(

2 3´

Solve quadratic for h 506

0 = h2 – 72

h = 2.93m So total distance = 7.13m (add on 4.2)

B

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.

e = 0.4m

T

2g

A

e 50T 21e gl= = =

lIn equilibrium

So length string = 1.4m in equilibrium

N

A At A PE = 2g1.4 KE = 0 EPE = 0

E PE = 0 KE = 2v2 EPE = 250 0.4

2 1´´

PE = -2g 0.2 KE = 0 EPE = 250 0.6)(

2 1´

0.2

0.4

initial acc. T – 2g = 2a

50 0.6 2 21 g a´ - =

subst x = 0.6 and l = 50

a = (500.6 - 20) = 5ms-2

B

N

A At A PE = 2g1.4 KE = 0 EPE = 0

E PE = 0 KE = 2v2 EPE = 250 0.42 1´´

PE = -2g(0.6) KE = 0 EPE = 250 0.6)(

2 1´

0.2

Energy at B = Energy at E

1 = v2

v = 1ms-1

0.4

So velocity at E = 1ms-1

B

-2g(0.6) + = 0 + 2v2 + 250 0.6

2 1´´

250 0.42 1´´

N = 1m

A

B = 1.6

At A PE = 2g1.4 KE = 0 EPE = 0

E = 1.4m PE = 0 KE = 2v2 EPE = 250 0.4

2 1´´

PE = -2g0.2 KE = 0 EPE = 250 0.6)(

2 1´

0.2

Energy at B = Energy at N

v =-3 impossible so it does not reach N

0.4PE = 2g0.4 KE = 2v2 EPE = 0

-2g(0.2) + = 2g0.4 + 2v2250 0.6

2 1´´

N

A

B

At A PE = 2g1.4 KE = 0 EPE = 0

PE = 2g(0.4-x) KE = 0 EPE =250 x

2 1´´

PE = -2g(0.2) KE = 0 EPE = 250 0.6)(

2 1´

0.2

Energy at B = Energy at M

x = 0.2m or 0.6mBut 0.6m is at the bottom so x = 0.2m

xM at M

-2g(0.2) + = 2g(0.4- x) + 250 0.6

2 1´´

250 x2 1´´

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

N

A

B

At A PE = 2g1.4 KE = 0 EPE = 0

0.2

Energy at B = Energy at M

x = 0.2m or 0.6mBut 0.6m is at the bottom so x = 0.2m

xM

-2g(0.2) + = 2g(0.4-x) + 250 0.6

2 1´´

250 x2 1´´

Finding the maximum height – M

So the length of the spring at the max height is 1.2m i.e 0.2m above E