Work and Energy

Card 1: What is work?

Work

Work done by a constant force is given by the product of the force and the distance moved in the direction of the force.

Unit: Nm or Joule (J)

Work is a scalar quantity.

Card 2: Formula of Work

Formula of work

Example 1

A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. Calculate the work being done by the force.

Answer:

Work done,
W = F × s × cos θ
W = 50 × 3.0 × cos30o = 129.9J

 

Card 3: Formula of work 2 - When the displacement is in the direction of force

Formula of work 2

When the direction of force and motion are same, θ = 0o, therefore cosθ = 1

Work done,

W = F × s

Example 2

Diagram above shows a 10N force is pulling a metal. The friction between the block and the floor is 5N. If the distance travelled by the metal block is 2m, find

  1. the work done by the pulling force
  2. the work done by the frictional force

Asnwer:

(a) The force is in the same direction of the motion. Work done by the pulling force,

      W = F × s = (10)(2) = 20J

(b) The force is not in the same direction of motion, work done by the frictional force

      W = F × s × cos180o= (5)(2)(-1) = -10J

 

Card 4: Work Done to Move Object Upward

Work Done Against the Force of Gravity

Example 3
Ranjit runs up a staircase of 35 steps. Each steps is 15cm in height. Given that Ranjit's mass is 45kg, find the work done by Ranjit to reach the top of the staircase.

Answer:

In this case, Ranjit does work to overcome the gravity.
Ranjit's mass = 45kg
Vertical height of the motion, h = 35 × 0.15
Gravitational field strength, g = 10 ms-2
Work done, W = ?

W = mgh = (45)(10)(35 × 0.15) = 2362.5J

 

Card 5: Finding Work Done from a Force - Displacement Graph

Force - Displacement Graph

In a Force-Displacement graph, work done is equal to the area in between the graph and the horizontal axis.

Example 4

The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m. Find the work done by the force to move the trolley.

Answer:

In a Force-Displacement graph, work done is equal to the area below the graph. Therefore, work done

            W = \frac{1}{2} \times 8 \times 10 = 40J
Card 6: What is Energy

Energy

Energy is defined as the capacity to do work.

Work is done when energy is converted from one form to another.

Unit: Nm or Joule(J)

Card 7: Kinetic Energy

Kinetic Energy

Kinetic energy is the energy of motion.

Example 5

Determine the kinetic energy of a 2000-kg bus that is moving with a speed of 35.0 m/s.

Answer:

Kinetic Energy,

            \begin{gathered}
  E_K  = \frac{1}
{2}mv^2  \hfill \\
  E_K  = \frac{1}
{2}(2000)(35)^2  \hfill \\
  E_K  = 1225000J \hfill \\ 
\end{gathered}  

Card 8: Gravitational Potential Energy

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object as the result of its vertical position (i.e., height).

Formula:

Example 6

A ball of 1kg mass is droppped from a height of 4m. What is the maximum kinetic energy possessed by the ball before it reached the ground?

Answer:

According to the principle of conservation of energy, the amount of potential energy losses is equal to the amount of kinetic energy gain.

Maximum kinetic energy
= Maximum potentila energy losses
= mgh = (1)(10)(4) = 40J

Card 9: Elastic Potential Energy

Elastic Potential Energy

Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing.

Formula:

Example 7

Diagram above shows a spring with a load of mass 0.5kg. The extention of the spring is 6cm, find the energy stored in the spring.

Answer:

The energy stored in the spring is the elestic potential energy.

            \begin{gathered}
  E_P  = \frac{1}
{2}Fx \hfill \\
  E_P  = \frac{1}
{2}(5)(0.06) = 0.15J \hfill \\ 
\end{gathered} 
            
Card 10: Conservation of Energy and Work Done

Conservation of Energy and Work Done

During a conversion of energy,

Amount of Work Done = Amount of Energy Converted

Example 8

A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and it stops at B after 4 seconds. What is the work done to overcome the friction?

Answer:

In this case, kinetic energy is converted into heat energy due to the friction. The work done to overcome the friction is equal to the amount of kinetic energy converted into heat energy, hence

              \begin{gathered}
  {\text{Work done}} \hfill \\
  {\text{ =  Kinetic energy losses}} \hfill \\
  {\text{ =  }}\frac{1}
{2}mv_1^2  - \frac{1}
{2}mv_2^2  \hfill \\
   = \frac{1}
{2}(5)(4)^2  - \frac{1}
{2}(5)(0) \hfill \\
   = 40J \hfill \\ 
\end{gathered} 
              

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