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Electrical energy, elastic potential energy

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Presentation on theme: "Electrical energy, elastic potential energy"— Presentation transcript:

1 Electrical energy, elastic potential energy
Efficiency of a small electric motor

2 Objectives To learn the energy changes that take place when a spring is stretched or compressed To be able to calculate the energy stored in a stretched spring To know how to calculate electrical energy Some examples of calculations involving energy transfers

3 The extension of this spring is
0.5m The spring stores……. Elastic potential energy

4 The EPE stored in a stretched spring is given by the formula
Elastic potential energy = ½ k x2 ‘X’ is the extension of the spring in metres ‘k’ is called the spring constant. It is measured in N/m. The stiffer the spring, the greater the value of ‘k’

5 At the top and bottom of the oscillation, the bob is not moving.
In the middle of the oscillation, the kinetic energy is a maximum

6 The total energy is constant

7 A spring, which has a spring constant k = 7
A spring, which has a spring constant k = 7.50 N/m, has been stretched 0.40 m from its equilibrium position. What is the potential energy now stored in the spring?

8  A spring with a spring constant k = 800 N/m has been compressed, and 196 J of potential energy is stored. What distance from equilibrium has the spring been compressed?

9 Electrical energy You should remember the formula for electrical power Power = current X voltage P = I X V You should also remember the relationship between power, energy and time. Power = energy ÷ time So an expression for electrical energy is: Electrical energy = current X voltage X time E = I X V X t

10 energy = current × voltage × time
For example, consider an LCD TV that is left on continuously for a week. It uses 230 V and 4 A. The energy transferred can be calculated using the equation: energy = current × voltage × time Teacher notes Students may need to be reminded that ‘M’ represents mega, i.e. million. Photo credit: © Levent Konuk, Shutterstock.com 2011 energy = 4 × 230 × (60 × 60 × 24 × 7) = 556,416,000 J = 556 MJ

11 200g of a liquid at 21⁰C was heated to 51⁰C by a current of 5A at 6V for 5 minutes. Neglecting heat losses, calculate the specific heat capacity of the liquid.

12 A 50W heating coil is totally immersed in 100g of water
A 50W heating coil is totally immersed in 100g of water. Assuming no heat energy is transferred to the surroundings: If the temperature of the water is 20⁰C when the heater is switched on, how long would it take for the water to boil? After the water has been boiling for 15 minutes it is found that the water in the flask has decreased to 80g. Calculate a value for the specific latent heat of vaporisation of the water. (specific heat capacity of water is 4200J/kg⁰C)

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