Energy and workSections 12, 13, 14 and 15

http://phet.colorado.edu/en/simulation/energy-skate-park Interactive website

Aims

Be able to use the equation Kinetic energy = ½ mV2

Be able to use the equation DPE = mgDH

Investigate and apply the conservation of energy

Be able to use the equation DW = FDS, including force not along the line of motion

Section 12 Energy symbol E unit JoulesEnergy is the ability to do work. Energy can take many forms: heat, electrical, kinetic, potential etc…

Whenever a mass moves it has kinetic energy:

Example:Calculate the kinetic energy of a car of mass 900 Kg travelling at 20 ms-1

Ek =0.5mV2 Ek = 0.5 x 900 x (20)2 Ek = 180 000 Joules

The equation for kinetic energy is:

Section 13 Change in gravitational potential energy

Example:A man of mass 70 kg climbs a ladder and moves through a vertical height of 4.5 m. Calculate the change in gravitational potential energy.

DEgrav = mgDh

DEgrav = 70 x 10 x 4.5

DEgrav = 3150 J

If a mass is lifted vertically in a gravitational field it gains gravitational potential energy

Section 15Work symbol W unit JoulesWork is the amount of energy converted from one form into another or the amount of energy used (e.g. kinetic in potential)

Whenever energy changes from one form into another work is done.

Mechanical work is given by:

Distance moved

Force applied

Example:A force of 20 N moves an object a distance of 300 m. Calculate the work done.

Work = force x distanceWork = 20 x 300Work = 6000 Joules

What happens to the energy (6000 J) if there is no friction?

All the energy is converted into kinetic energy. Hence K.E. = 6000 J

What happens to the energy (6000 J) if there is friction?

Only some of the energy is kinetic energy the rest is converted into heat, sound etc.

The force and the distance have to be parallel for the equation to work.If they are not then you need to find the component of the force parallel to the direction of movement.

Force applied

Force F

Force F can be replaced by two components

q One vertically:F x Sin q

The other horizontally:F x Cos q

Direction of movement

F x Cos q

Distance moved d

Work done by force F is: Work = F x Cos q x d

Example:

If F = 200 N (q = 40o ) moves the object horizontally a distance of 50 m, calculate the work done by the force F

Force F

q

Work = F x Cos q x dHence Work = 200 x Cos 40 x 50

Work = 7660 Joules

Section 14Conservation of energyEnergy can neither be created nor destroyed. It can only be converted from one form into another. For most systems ignoring friction:

A

B

C

If the yellow object moves from A to B to C the total energy is fixed.

Throughout the journey the sum of kinetic energy and potential energy is a constant

Work is the amount of energy converted from one form to another.

If friction acts then the work done by friction can be calculated using:force x distance

The previous equation is now altered to take the frictional force into account and is rewritten as:

ExampleA stone is thrown upwards with a kinetic energy of 200 J, it rises to a height (24 m) such that it gains 140 J of gravitational potential energy. How much work is done and what is the size of the frictional force.

Using the above equation:KEb + PEb = KEa + PEa + work done

200 + 0 = 0 + 140 + Work doneHenceWork done = 60 Joules

Work done = Force x distanceThe distance is 24 m and the work done is 60 J

Hence Force = 60/24 = 2.5 Newton

Be able to use Kinetic energy = ½ mV2

Calculate the kinetic energy of a car (1200 kg) moving with a speed of 20 m s-1.

Answer: 240 000 Joules

Be able to use the equation DPE = mgDH

Elevations are in metres and g = 9.81 m s-2 A ball of mass 0.10 Kg rolls down the hill. Calculate the change in gravitational potential energy.Answer: -12.753 Joules

Be able to use the equation DW = FDS (example 1)

The tugboat pulls on the ship with a force of 30 000 N, it maintains this force over a distance of 400 metres. Calculate the work done.Answer: 12 000 000 Joules

Two tug boats each providing a force of 20 000 N are pulling a ship. The angle between the direction of movement and the ropes is 20o in each case. Calculate the work done by each tugboat if the ship is pulled a distance of 1200 metres.Answer: 22552623 Joules.

Be able to use the equation DW = FDS (example 2)

Apply the conservation of energy

A man of mass 80 Kg is placed in the cradle and given a velocity of 15 m s-1 upwards. If there is no friction what height will he reach? Answer: 11.46 mIf he reaches a height of 9.5 metres, what was the size of the frictional force? Answer: (-) 162 Newton