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SS2: PHYSICS - 1ST TERM

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  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
    |
    1 Quiz
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
    |
    1 Quiz
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz



Topic Content:

  • Meaning/Types of Energy
  • Meaning/Examples of Machines
  • Terms Associated with Machines

Energy is the ability to do work. There are two types of energy that constitute mechanical energy, they are Potential energy and Kinetic energy.

Mechanical energy is the sum of the kinetic energy and potential energy of an object that is used to do a particular work. In other words, it describes the energy of an object because of its motion, position, or both.

Potential energy, P.E = mgh (Energy of a body by virtue of its position or rest or stored energy due to height).

m = mass of the body

g = acceleration due to gravity

h = height to which the object is raised

Kinetic energy, K.E =  \( \frac{1}{2} \scriptsize mV^2 \) (Energy of  body in motion)

m = mass of the body

V = Velocity of the body

The principle of the conservation of mechanical energy states that the total mechanical energy in a system (i.e., the sum of the potential plus kinetic energies) can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature.

The law of conservation of mechanical energy shows that P.E + K.E = constant.

Machines:

A machine is a device or a contrivance with which force is applied at one end (Effort) to overcome load (L) at the other end.

A small force (effort) can be applied to overcome a larger load.

Examples of such machines are levers, pulleys, inclined planes, screws, hydraulic presses, etc.

Terms Associated with Machines:

Let’s define some terms that apply to the workings of machines.

Mechanical Advantage of Force Ratio:

This is the ratio of load to effort. It can be defined as the ability of a machine to overcome a large load through a small effort.

M.A =  \( \frac{Load}{Effort} \\= \frac{L}{E} \)

It has no unit and is also referred to as the Force Ratio.

Since load is the output force and effort is the input force, Mechanical Advantage can also be defined as:

M.A =  \( \frac{Output\:force}{Input\:force} \)

Velocity Ratio:

This is the ratio of distance moved by effort to the distance moved by load.

V.R = \( \frac{distance \: moved \: by \: effort}{distance \: moved \: by \: load} =\frac{x}{y} \)

When load is greater than effort, M.A > 1, V.R > 1.

Efficiency:

This is the ratio of work output to the work input of a machine expressed as a percentage.

Efficiency = \( \frac{Work \:output}{Work\: input} \scriptsize \: \times \: 100 \%\)

But work = \( \scriptsize Force \: \times \: distance\: / \: displacement \)

E = \( \frac{Load \: \times \: distance \: moved \: by \: load}{Effort \: \times \: distance \: moved \: by \: effort} \scriptsize \: \times \: 100 \% \)

E = \( \frac{L}{E} \: \times \: \frac{y}{x} \: \times \: \scriptsize 100 \% \)

E = \( \frac{\normalsize\frac{L}{E}}{ \normalsize\frac{x}{y}}\scriptsize \: \times \: 100 \% \)

E = \( \frac{M.A}{V.R}\scriptsize \: \times \: 100 \% \)

For an ideal machine, the efficiency is 100%, but due to friction, mechanical advantage (M.A) decreases, and this also affects (reduces) the efficiency of the machine.

Example 1.1.1:

A machine has an efficiency of 60%. If the machine is required to overcome a load of 30 N with a force of 20 N, Calculate its Mechanical Advantage.

Solution

Load = 30 N
Effort = 20 N

Use the formula:

Mechanical Advantage, M.A =  \( \frac{Load}{Effort} \)

M.A =  \( \frac{30\:N}{20\: N}\\ = \scriptsize 1.5 \)

Example 1.1.2:

The velocity ratio of a machine is 5 and its efficiency is 75%. What effort would be needed to lift a load of 150N with the machine?

Solution

Velocity Ratio (V.R) = 5
Efficiency = 75% or \( \frac{75}{100} \)
Load = 150 N

First, let’s calculate the Mechanical Advantage (M.A):

E = \( \frac{M.A}{V.R}\scriptsize \: \times \: 100 \% \)

⇒ \(\frac{75}{100} = \frac{M.A}{5} \)

M.A = \(\frac{75\:\times \:5} {100} \)

M.A = 3.75

Since we know the load and the mechanical advantage, we can use the formula below to calculate the effort.

M.A =  \( \frac{Load}{Effort} \)

3.75 =  \( \frac{150}{Effort} \)

Effort = \( \frac{150}{3.75} \)

Effort = 40 N

Example 1.1.3:

A machine with velocity ratio of 5 requires 1200 J of work to lift a load of 400 N through a vertical distance of 2.0 m. Calculate the efficiency and mechanical advantage of the machine.

Solution

Efficiency = \( \frac{Work \:output}{Work\: input} \scriptsize \: \times \: 100 \%\)

Work input = 1200 J

Work output = \( \scriptsize Force \: \times \: distance \)

∴ Efficiency = \( \frac {400 \: \times \: 2}{1200} \scriptsize \: \times \: 100\% \\ = \frac {800}{1200} \scriptsize \: \times \: 100\% \\ \scriptsize = 66.7 \%\)

Since we now have a value for efficiency, we can use the formula below to calculate mechanical advantage.

E = \( \frac{M.A}{V.R}\scriptsize \: \times \: 100 \% \)

\( \frac {66.7}{100} = \frac {M.A}{5} \)

M.A = \( \frac {66.7\: \times \:5}{100} \)

= 3.33.

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