Energy is transferred or transformed from one energy form to another via an energy pathway. In this post, we will briefly discuss the following topics: (a) different forms of energy (b) different mechanisms of energy transformation or conversion or transfer (c) Numerical problems related to energy conversion or transformation with the solution.
Here let’s quickly revise some fundamentals related to energy transformation. This will help us to solve the numerical problems.
Different Energy forms or stores include:
• elastic and magnetic
Energy pathways (or energy transfer mechanisms) include:
• electric (a charge moving through a potential difference)
• mechanical (a force acting through a distance)
• heating (driven by a temperature difference)
• waves (such as electromagnetic radiation or sound waves).
The rule of conservation of energy is never broken.
In mechanical systems, the energy transferred between stores (such as gravitational to kinetic) is equivalent to the work done.
The equation of work done = force × distance applies when the force is constant. However, if the force is not constant then to calculate the work done you need to calculate the area under the force-distance graph.
Numerical Problems on Energy transformation or conversion with solution
Here we have listed and solved 4 numerical problems (based on energy conversion).
Numerical Problem 1
A boy drags a box to the right across a rough, horizontal surface using a rope that pulls upwards at 25° to the horizontal.
a) Once the load is moving at a steady speed, the average horizontal frictional force acting on the load is 470N.
Calculate the average value of F
b) The load is moved by a horizontal distance of 250m in 320 s.
Calculate the work done on the load by F
a) The frictional force acts to the left. The horizontal force to the right must equal 470N for the resultant force to be zero with no
change in velocity.
Resolving F cos25 = 470 gives F = 518N
b) The work done = force × distance = 518 × 250 = 130 kJ.
Numerical Problem 2
A diver climbs to a diving board and dives from it.
The height of the diving board above the floor is 4.0m. The mass of the diver is 54kg.
a) Calculate the gain in gravitational potential energy when the diver climbs to the diving board.
b) The diver enters the water at a speed of 8.0m/s. Calculate the kinetic energy of the diver as she enters the water.
c) Suggest why the kinetic energy of the diver in part (b) is different from the gravitational potential energy gained in part (a).
a) gain in gravitational potential energy = ΔE = mgh = 54 × 9.81× 4.0 = 2.1 kJ
b) ΔEk =(1/2) m(v^2 − u^2 )
Here u = 0 as the diver starts from rest.
ΔEk = (1/2)mv^2 = (1/2)x54x8^2 =1.7 kJ
c) A number of factors can be discussed here.
• Work is done against air resistance.
• The distance travelled by the centre of mass of the diver falling to the water is not equal to the distance gained in climbing the
• The diver gains gravitational potential energy in taking off.
• Energy usually goes into rotational kinetic energy while diving.
formulas of power and using those to solve numerical problems
Energy can be transferred at different rates. Think of two boys of equal weight who run up a hill: the quicker boy is the more powerful of the two.
As work done = force × distance,
power = force × distance/time
or Power= force × distance/time
Therefore power = force × speed. The area under a force–speed graph gives the power developed during an energy transfer.
Numerical Problem 3
Water in a hydroelectric system falls vertically to a river below at the rate of 12 000kg every minute.
The water takes 2.0 s to fall this distance. It has zero velocity at the top.
a) Calculate the height through which the water falls.
b) A small electrical generator of efficiency 20% is at the foot of the system. All the water goes through the generator.
Determine the electrical power output of the generator.
c) Outline the energy transfers in this system.
a) The acceleration, g, is uniform, so suvat equations can be used.
s = ut +(1/2) g t^2 = 0 + (1/2)x9.81×2^2
=> s = 20 m
b) From this point, it is best to work in seconds rather than minutes.
Mass of water flowing every second = 12000/60 = 200 kg
Gravitational potential energy transferred every second = mgh = 200 × 9.81× 19.8 = 38847.6 J
As the generator is 20%, the power output is 38847.6/5 = 7769.52W.
c) The water has stored gravitational potential energy at the top of the waterfall. As the water falls, energy is transferred into kinetic
energy. At the bottom of the waterfall, the maximum transfer of energy has occurred. The water enters a turbine where the linear
kinetic energy of the water is transferred into the rotational kinetic energy of the turbine and the dynamo. The dynamo converts this
kinetic energy into electrical energy, wasted thermal energy, and frictional losses.
Numerical Problem 4
Q)An electric motor pulls a glider horizontally from rest to a constant speed of 27.0m/s with a force of 1370N in a time of 11.0s. The motor has an overall efficiency of 23.0%.
Determine the average power input to the motor.
Avg Speed = 148.5/11 = 13.5 m/s
Power output = force x velocity = 1370 x 13.5 = 18495 W
efficiency is 23%.
So Avg power input = (power output/23) x 100 80.4 kW