How Do You Find The Kinetic Energy

What is Kinetic Energy

Kinetic Energy: The energy of a body due to its motion is called kinetic energy. In other words. The ability of a body to do work by virtue of its motion is called its kinetic energy.

Expression for Kinetic Energy: The kinetic energy of a body is measured in terms of the amount of work done by an opposing force that brings the body to rest from its present state of motion.
Kinetic-Energy
Suppose a body of mass m is moving with a velocity v and is brought to rest by an opposing force F.
Now retarding force is given by
F = ma          …(1)
Now using the equation of motion,
v2 – u2 = 2as, we get
02 = v2 – 2as
∴s=\( \frac{{{v}^{2}}}{2a} \) ….. (2)
Kinetic energy of the body = work done by the retarding force
or Kinetic energy = force × displacement
= F . s …(3)
Substituting the value of F from equation (1) and the value of s from equation (2) in equation (3), we get
\( K.E.=ma\times \frac{{{v}^{2}}}{2a} \)
\( K.E.=\frac{1}{2}m{{v}^{2}} \)
Thus, a body of mass m and moving with a velocity v has the capacity of doing work equal to \(\frac { 1 }{ 2 }\) mv2 before it stops.

Kinetic Energy Example Problems With Solutions

Example 1:    A bullet of mass 100 gm is fired with a velocity 50 m/s from a gun. Calculate the kinetic energy of the bullet.
Solution:    Kinetic energy is given by
\( K.E.=\frac{1}{2}m{{v}^{2}} \)
Here m = 100 gm = 0.1 kg; v = 500 m/s
K.E. = \(\frac { 1 }{ 2 }\) × 0.1 × (50)2
= \(\frac { 1 }{ 2 }\) × 0.1 × 50 × 50 = 125 J

Example 2:    A 4 kg body is dropped from the top of a building of height 2.5 m. With what velocity will it strike the ground ? What is its kinetic energy when it strikes the ground ?
(Takes g = 9.8 m/s2)
Solution:    Velocity of the body with which it strikes the ground can be calculated by using the equation, v2 = u2 + 2gh
Here u = 0; g = 9.8 m/s2 ; h = 2.5 m
Substituting these values, we get
v2 = 02 + 2 × 9.8 × 2.5 = 49
or v = 7 m/s
Thus, the speed of the body with which it strikes the ground = 7 m/s.

Example 3:   Calculate the velocity of 4 kg mass with kinetic energy of 128 J.
Solution:    The formula for kinetic energy is given by
\( K.E.=\frac{1}{2}m{{v}^{2}} \)
Here K.E. = 128 J; m = 4 kg
128 = \(\frac { 1 }{ 2 }\) × 4 × v2
v2 = 64
v = 8 m/s

Example 4:    Which would have a greater effect on the kinetic energy of an object, doubling the mass or doubling the velocity ?
Solution:    (i) The kinetic energy of a body is directly proportional to its “mass” (m). So, if we double the mass (so that it becomes 2m), then the kinetic energy will also get doubled.
(ii) On the other hand, kinetic energy of a body is directly proportional to the “square of its velocity” (v2). So, if we double the velocity (so that it becomes 2v), then the kinetic energy will become four times. This is because : (2v)2 = 4v2.
It is clear from the above discussion that doubling the velocity has a greater effect on the kinetic energy of an object.

Example 5:    Two bodies of equal masses move with uniform velocity v and 3v respectively. Find the ratio of their kinetic energies.
Solution:    In this problem, the masses of the bodies are equal, so let the mass of each body be m. We will now write down the expression for the kinetic energies of both the bodies separately.
(i) Mass of first body = m
Velocity of first body = v
So, K.E. of first body = \(\frac { 1 }{ 2 }\) mv2         ….. (1)
(ii) Mass of second body = m
Velocity of second body = 3v
So, K.E. of second body = \(\frac { 1 }{ 2 }\) m (3v)2
= \(\frac { 1 }{ 2 }\) m × 9 v2
= \(\frac { 9 }{ 2 }\) mv2                 ….. (2)
Now, to find out the ratio of kinetic energies of the two bodies, we should divide equation (1) by equation (2), so that
\( \frac{\text{K}\text{.E}\text{.}\,\text{of}\,\text{first}\,\text{body}}{\text{K}\text{.E}\text{.}\,\text{of}\,\text{sec}\,\text{ondbody}\,}=\frac{\frac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}}{\frac{\text{9}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}} \)
\( \frac{\text{K}\text{.E}\text{.}\,\text{of}\,\text{first}\,\text{body}}{\text{K}\text{.E}\text{.}\,\text{of}\,\text{sec}\,\text{ond}\,\text{body}}=\frac{\text{1}}{\text{9}}\text{ }….\text{ (3)} \)
Thus, the ratio of the kinetic energies is 1 : 9. We can also write down the equation (3) as follows:
K.E. of second body = 9 × K.E. of first body
That is, the kinetic energy of second body is 9 times the kinetic energy of the first body. It is clear from this example that when the velocity (or speed) of a body is “tripled” (from v to 3v), then its kinetic energy becomes “nine times”.

Example 6.  What is the kinetic energy of a 1200 kg car travelling at a velocity of 25 m s-1?
Solution:
How Do You Find The Kinetic Energy

Example 7. Table shows the masses and velocities of two boys, Bek Sar and Kek Cik respectively.
How Do You Find The Kinetic Energy 1
Which of the boys has greater kinetic energy?
Solution:
How Do You Find The Kinetic Energy 2

How to Calculate the Energy Used

What is Energy

Definition: Energy is the ability to do work. The amount of energy possessed by a body is equal to the amount of work it can do when its energy is released. Thus, energy is defined as the capacity of doing work. Energy is a scalar quantity and it exists in various forms.

Units of energy: The units of energy are the same as that of work. In SI system, the unit of energy is joule (J). In CGS system, the unit of energy is erg.

  • 1 Joule = 107 ergs
  • Other units of energy in common use are watt-hour and kilowatt hour.
  • 1 watt-hour = 1 watt × 1 hour
    = 1 watt × 60 × 60 sec
    = 3600 J
  • 1 kilowatt-hour (kWh) = 3.6 × 106 Joule
  • Heat energy is usually measured in calorie or kilocalorie such that
    1 calorie = 4.18 J
  • A very small unit of energy is electron volt(eV).
    1 eV = 1.6 × 10-19 J

The energy possessed by a body due to its state of rest or state of motion is called mechanical energy.
Mechanical energy is of two types
(A) Kinetic Energy            (B) Potential Energy.

Principle of Conservation of Energy

  1. Principle of conservation of energy states that energy cannot be created or destroyed but can be changed from one form to another.
  2. The total energy of the universe is constant. The total energy of an isolated system is constant.
  3. Figure shows the transformation of energy from one form to another.
    Principle of Conservation of EnergyPrinciple of Conservation of Energy 1

Principle of Conservation of Energy Experiment

Aim: To study the principle of conservation of energy.
Materials: Ticker tape, a polystyrene sheet, string, a 300 g slotted mass, a pulley, cellophane tape
Apparatus: Ticker timer, a.c. power supply, two retort stands with clamps, trolley, electronic balance, plane
Method:

  1. A friction-compensated plane is arranged as shown in Figure.
  2. The mass of the trolley, m1 is measured with an electronic balance.
  3. A slotted mass of mass, m2 = 300 g is tied to one end of a non-elastic string.
  4. The other end of the string is tied to one end of the trolley.
  5. The string is placed over the pulley and held at 0.5 m above the polystyrene sheet.
  6. The ticker timer is switched on and the slotted mass is released so that it falls downward, pulling the trolley down the runway.
  7. The ticker timer is analysed to determine the final velocity of the trolley.
    Principle of Conservation of Energy 2

Results:
Principle of Conservation of Energy 3
Discussion:

  1. The plane is friction-compensated to minimise energy loss due to friction.
  2. When the slotted mass drops, it loses gravitational potential energy. The trolley and slotted mass gain kinetic energy.
  3. Ideally the gravitational potential energy loss, Ep equals to the kinetic energy gain, Ek. This is in accordance to the principle of conservation of energy.
  4. However, the experimental results shows that Ek is slightly less than Ep. This is because of unavoidable energy loss due to friction of the trolley as well as air friction.