Work and energy are closely related concepts that explain how force and motion are connected. Work is done when a force acts on an object and causes it to move through a certain distance. Energy, on the other hand, is the ability or capacity to do work. Both are scalar quantities and are measured in joules (J).
Work
Work is a form of energy transfer that occurs when a force moves an object through a certain distance in the direction of the force. For work to be done, both force and displacement are necessary. The concept of work was introduced by the French mathematician Gaspard Coriolis in 1826. The SI unit of work is joule (J).
Formulas:
\begin{matrix} W = \bigtriangleup KE \\ KE = \frac{1}{2} mv^{2} \\ W = F \bigtriangleup r \cos \Theta \end{matrix}
Energy
Energy can be defined as the measurement of the ability of something to do work. It is not a material substance. Energy can be stored and measured in many forms. It is also referred to as the force that works at a certain distance. Energy deals with the capacity of an object to do the work. The SI unit of physics is joules
Formula
E_{k} = \frac{1}{2} mv^{2} Here,
Ek = kinetic energy of object
m = mass of object
v = speed of object
Below is a table of differences between Work and Energy
Work | Energy |
|---|---|
| Work is the ability to supply force and a change in distance to an object. | Energy is the ability to supply or create work. |
| There is a parallel relationship between the force components and displacement | Energy is the result of the work performed |
| The action did on the thing causing some displacement | It is described as a property of a system |
| Scalar units. | Scalar units. |
| Work = force × distance | There are various equations depending upon the kinds of energy |
| If the applied force is within the same direction of the displacement then work is positive | there is no direction component here because it is a scalar quantity |
| If the applied force is within the other way of the displacement that employment is negative | Then also there'll be no direction component here because it may be a scalar quantity |
| Work was only utilized in 1826 | Energy was coined in 4 BC |
Sample Problems
Question 1: If a force of 30 N in lifting a load of 2kg to a height of 10m (g = 10ms-2), then calculate the amount of work done in this process?
Solution: Given, Force in lifting mg = 30 N; height = 10 m
W = F.S (or) mgh
= 30 × 10
W = 300 J
Hence the answer is 300J
Question 2: Compute the work done if 10 N of force acts on the body showing the displacement of 2 m?
Solution: Given, F (Force) = 10 N,
d (Displacement) = 2 m,
W (Work done) = F × d
= 10 N × 2 m
= 20 Nm.
Question 3: A body of mass 10kg at rest is subjected to a force of 16N. Find the K.E. at the top of 10 s.
Solution: Mass m = 10 kg
Force F = 16 N
time t = 10 s
a = F/m
we know v = u + at
Kinetic energy KE: 1/2mv2
0.5 × 10 × 16 × 16
1280J
Question 4: A body of mass 5kg is thrown up vertically with a K.E. of 1000 J. If acceleration thanks to gravity is 10ms-2, find the peak at which the K.E. becomes half the first value.
Solution: Mass m = 5kg
K.E Energy = 1000J
g = 10m s-2
At a height 'h', mgh = E/2
5 × 10 × h = 1000/2
h = 500/50
h = 10m
Unsolved Problems
Question 1: A force of 35 N lifts a body of mass 3 kg to a height of 10 m (g = 10 m/s²). Find the work done.
Question 2: A body of mass 6 kg initially at rest is acted upon by a force of 18 N for 4 seconds. Find the kinetic energy gained.
Question 3: A body of mass 5 kg is thrown upward with a kinetic energy of 500 J. Find the height at which its kinetic energy becomes zero.
Question 4: A force of 25 N acts on a body of mass 5 kg initially at rest for 6 seconds. Calculate the total kinetic energy gained by the body.
Question 5: A body of mass 3 kg is thrown vertically upward with a velocity of 20 m/s. Calculate the kinetic energy of the body at a height of 15 m (g = 10 m/s²).