# Physics: question and answer on elasticity and Hooke’s law 1

Elastic wire extended by 1.0cm when a load of 20g hangs from it. What additional load will be required to cause further extension of 2.0cm?

1.0cm = 0.01m.  2.0cm = 0.02m

The force exerted by 20g

Convert the 20g to kg  and multiple it by 10

$=\quad \frac { 20 }{ 1000 } \times 10\quad =0.2N$

Using Hooke’s law,  F= ke

Therefore  0.2 = 0.01k

k = 20N/m

let m = the additional mass so

(0.2 +M)10 = 20 x (0.01 +0.02)

0.2 =10M = 20 x 0.03 = 0.6

10M = 0.4

M = 0.04kg

# Question and answer on elasticity and Hooke’s law 2

State Hooke’s law

Hooke’s law state that the extension e of elastic materials is directly proportional to the load or force applied provided the elastic limit is not exceeded. F ∝e or F = ke, where k = elastic constant or stiffness.

# Question and answer on elasticity and Hooke’s law 3

Define the following in elasticity of solid

(a). Elastic limit, (b) Yield point and (c) Breaking point

(a). Elastic limit is the maximum extension of elastic material by the force applied beyond which the material would not return to its original length or lost its elastic property .

(b). Yield point, this is the point at which the material has completely lost its elastic property, and instead of elastic it becomes plastic.

(c). Breaking point: This the point at which the material cannot stand any further increase in a load. At this point the material breaks when further load is applied.

# Question and answer on elasticity and Hooke’s law 4

Define Elasticity of a solid

Elasticity of a solid can be defined as the ability of the solid to return to its original length or shape when stretched or compressed and released

# question and answer on elasticity and Hooke’s law 5

A spiral spring is compressed by 0.02m, calculate the energy stored in the spring, if the force constant is 400N/m

Using  W = ½ke2, where k = 400N/m, and e = 0.02

So W = ½ x 400 x (0.02)2

W = 0.5 x 400 x 0.0004 =0.08J

Energy stored in the spring = 0.08J

# question and answer on elasticity and Hooke’s law 6

A stone of mass 20 g is released from a catapult whose rubber has been stretched through 4cm. If the force constant of the rubber is 200N/m. calculate the velocity with which the stone leaves the catapult.

Using ½ke2 = ½mv2, where K = 200N/m, e = 4.0cm = 0.04m, m = 20g =0.02kg

½ x 200 x (0.04)2 =  ½ x 0.02 x v2

0.5 x 200 x 0.0016 = 0.5 x 0.02 v2

o.16 =0.01v2

V2 =16/0.01  = 16

V2  = 16

V =4m/s

# Question and answer on elasticity and Hooke’s law 7

A spring 20cm long is stretched to 25cm by a load of 50N, what will be its length when stretched by 100N, assuming that the elastic limit is not reached.

F = ke , where e = 5cm = 0.05m,  F = 50N

So F = ke

50 =0.05k

K =1000N/m

When force of 100N is applied

The F = ke

= 100 = 1000 x e

100 = 1000e

e = 100/1000= 0.1m

e = 0.1m = 10cm

so the length of the spring when 100N is applied = 30cm

# Question and answer on elasticity and Hooke’s law 8

A spring of constant force 1500N/m is acted upon by a constant force of 75N. Calculate the potential energy stored in the spring.

First find the extension using  F = ke, where k =1500N/m, and F =75N

So e = 75/1500 = 0.05m

Using w = ½ke2

W = ½ x 1500 x (0.05)2

W = 0.5 x 1500 x  0.0025  = 1.9J

# Question and answer on elasticity and Hooke’s law 9

A string of natural 1.5cm is extended 0.0050m by 0.8N, what will its length be when the applied force is 3.2N

F = Ke, where F = 0.8N, e = 0.0050m

0.8 = 0.0050k

K = 0.8/0.0050 = 160N/m

When F = 3.2N,

3.2 = 160e

e = 3.2/160 = 0.02m

e =0.02m = 0.2cm

Therefore the length when force of 3.2N is applied = 0.2 + 1.5 = 1.7cm

# question and answer on elasticity and Hooke’s law 10

A Spring balance designed for maximum load of 400N, was used to weigh a load of 500N. When the 500N load was removed, the pointer could not return to zero on the scale because

(a). the spring balance of a cheap type

(b). Elastic limit of the material has been exceeded

(c). Pointer was not properly adjusted

# Question and answer on elasticity and Hooke’s law 11

A catapult is used to project a stone. Which of the following Energy conversions takes place as the stone is released.

(a).the Kinetic energy of the stone is converted into gravitational potential energy

(b). The gravitational potential is converted into kinetic energy of the stone

(c). the elastic potential energy of the catapult is converted into kinetic energy of the stone

(d). the gravitational potential energy is converted to elastic energy.

# Question and answer on elasticity and Hooke’s law 12

(3) Define young modulus

The young’s modulus E, can be defined as :

$E\quad =\frac { Tensile\quad Stress }{ Tensile\quad Strain } \\ \\ \\ where\quad Stress\quad =\frac { Force }{ Area } \quad and\\ \\ Strain\quad =\quad \frac { Extension\quad e }{ Load }$

# Question and answer on elasticity and Hooke’s law 13

What is elastic potential Energy?

Elastic potential energy can be defined the energy possessed by an elastic material due its position. It may be compressed or stretched to possess such.  It can also be defined as work done in compressing of stretching elastic material. And is give as

W =½Fe or w = ½ke2 since F =ke

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