Back to Course

SS1: PHYSICS – 3RD TERM

0% Complete
0/0 Steps
  1. Production of Electric Current | Week 1
    6 Topics
    |
    1 Quiz
  2. Electric Current | Week 2
    5 Topics
    |
    1 Quiz
  3. Electrical Resistance of a Conductor | Week 3
    5 Topics
    |
    1 Quiz
  4. Particulate Nature of Matter | Week 4
    5 Topics
    |
    1 Quiz
  5. Crystalline and Non-crystalline Substances | Week 5
    3 Topics
    |
    1 Quiz
  6. Elastic Properties of Solids | Week 6 & 7
    4 Topics
    |
    1 Quiz
  7. Fluids at Rest & in Motion | Week 8 & 9
    6 Topics
    |
    1 Quiz
  8. Solar Collector
    3 Topics
    |
    1 Quiz



  • Do you like this content?

  • Follow us

Lesson Progress
0% Complete

Work Done in Springs and Elastic Materials:

When elastic materials are compressed or stretched, work is done. If the applied force is F and the extension produced is e, then,

The work done = average Force x extension

⇒ \( \scriptsize Average\:force = \normalsize \frac{initial\: force \: + \: final \: force}{2} \)

= \(\left( \frac {0 \:+\: F}{2} \right) \scriptsize \times e \)

⇒ \(\scriptsize Average\: Work\: done = \normalsize \frac{1}{2}\scriptsize Fe \)

But from Hooke’s law,

F = k.e

∴  \(\scriptsize Work\: done = \normalsize \frac{1}{2}\scriptsize ke.e \)

= \( \frac{1}{2}\scriptsize ke^2 \)

k = elastic constant, e = extension or compression produced 

Work done is in Joules.

Example 1:

Calculate the work done when a force of 20N stretches a spring by 50mm.

Solution:

Force = 20N
extension, e = \( \left(\frac{50}{1000} \right)\scriptsize m \\ = \scriptsize 0.05m\)

⇒ \(\scriptsize Work\: done = \normalsize \frac{1}{2}\scriptsize Fe \\ = \normalsize \frac{1}{2}\scriptsize \: \times \: 20 \: \times \: 0.05 \\ = \scriptsize0.5 \: J \)

Example 2:

If the work done in stretching a material is 375J, calculate the extension of the material if the force applied was 15N. What is the stiffness of the material?

Solution:

Work done = 375J
Force = 15N
e = ?
k = ?

⇒ \(\scriptsize Work\: done = \normalsize \frac{1}{2}\scriptsize Fe \)

⇒ \(\scriptsize 375J = \normalsize \frac{1}{2}\scriptsize \: \times \: 15N \: \times \: e \)

Make e the subject of the formula;

⇒ \( \scriptsize e = \normalsize \frac{375 \: \times \: 2}{15} \\ = \normalsize \frac{750}{15} \\ = \scriptsize 50m \)

To calculate the stiffness k, we will use the formula

F = ke

k = \( \frac{F}{e} \)

k = \( \frac{375}{50} \)

k = 7.5 Nm-1

Elastic Potential Energy:

Screenshot 2022 05 27 at 00.55.39

The energy that is stored in a compressed elastic material or stretched elastic material is equal to the work done in compressing or stretching the material. The type of energy that is stored in the material is called elastic potential energy because it is potentially able to do work.

The ability of a stretched or compressed elastic material to do work is called elastic potential energy.

⇒ \(\scriptsize Elastic\: Potential\: Energy, \: W = \normalsize \frac{1}{2}\scriptsize Fe = \normalsize \frac{1}{2}\scriptsize ke^2 \)

Elastic potential energy is stored energy that can be transformed into other forms of energy. For example, this happens when a catapult is used and an object is projected from it. According to the law of conservation of energy, the elastic potential energy stored in the rubber is transformed into the kinetic energy of the flying object.

elastic potential energy

Example 3:

Calculate the energy stored in a stretched spring with elastic constant 10000Nm-1, when a force of 200N is applied to its end?

Solution:

F = 200N
k = 10000Nm-1
e = ?

First we calculate extension, e using the formula:

F = ke

e = \( \frac{F}{K} \)

e = \( \frac{200}{10000} \)

e = 0.02m

⇒ \(\scriptsize W = \normalsize \frac{1}{2}\scriptsize Fe = \normalsize \frac{1}{2}\scriptsize ke^2 \)

We can use both formulas to get the same result:

using ⇒ \(\scriptsize W = \normalsize \frac{1}{2}\scriptsize ke^2 \)

⇒ \(\scriptsize W = \normalsize \frac{1}{2}\scriptsize \: \times \: 10000 \: \times \: 0.02^2 \)

⇒ \(\scriptsize W = \scriptsize 5000 \: \times \: 0.0004 \)

⇒ \(\scriptsize W = 2J \)

using ⇒ \(\scriptsize W = \normalsize \frac{1}{2}\scriptsize Fe \)

⇒ \(\scriptsize W = \normalsize \frac{1}{2}\scriptsize \: \times \: 200 \: \times \: 0.02 \)

⇒ \(\scriptsize W = \scriptsize 100 \: \times \: 0.02 \)

⇒ \(\scriptsize W = 2J \)

Example 4:

A stone of mass of 30 g was released by a catapult which was stretched through 5cm. If the force constant of the catapult was 175Nm-1, calculate the velocity with which the stone left the catapult.

Solution:

Let the velocity of the stone be v ms-1
e = \( \scriptsize 5cm \\ = \left( \frac{5}{100} \right) \scriptsize m \\ = \scriptsize 0.05m\)

k = 175Nm-1
m = \( \scriptsize 30g \\ = \left( \frac{30}{1000} \right) \scriptsize kg \\ = \scriptsize 0.03kg\)

The elastic potential energy of the stretched rubber of the catapult is transformed into kinetic energy of the stone.

Therefore,

⇒ \(\scriptsize \normalsize \frac{1}{2}\scriptsize ke^2 = \frac{1}{2}\scriptsize mv^2 \)

⇒ \( \normalsize \frac{1}{2} \scriptsize \: \times \: 175 \: \times \: 0.05^2 = \normalsize \frac{1}{2} \scriptsize \: \times \: 0.03 \: \times \: v^2 \)

⇒ \( \scriptsize 0.015 \: \times \: v^2 = \normalsize \frac{1}{2} \scriptsize \: \times \: 0.4375 \)

⇒ \( \scriptsize 0.015 \: \times \: v^2 = \scriptsize 0.21875 \)

⇒ \( \scriptsize v^2 = \normalsize \frac{0.21875}{0.015} \)

⇒ \( \scriptsize v^2 = 14.58 \)

⇒ \( \scriptsize v = \sqrt{14.58} \)

⇒ \( \scriptsize v = 3.83 \: ms^{-1} \)

Responses

Your email address will not be published. Required fields are marked *

error: