1. Motion on an Incline Energy solution
The principle states that energy can neither be created or destroyed.
Energy can only be transferred from one form to another.
If the transfer of energy is directly from one form into another the transfer is said to be 100% efficient.
Many engineering systems require the transfer of potential energy (PE) into kinetic energy (KE), or KE into PE
If the system is perfectly efficient (no losses) the two forms will equate:-
Potential energy at start = Kinetic energy at end
mgh = ½ mv2
JR/2008

2. Motion on an Incline Energy solution
If losses are present – typically in the form of friction then:-
Potential energy at start = Kinetic energy at end + work done to overcome friction
mgh = ½ mv Friction force x distance moved against friction force
(Rem:- work done = energy used = force x distance)
Note:-If a combination of forces act against motion (friction, air resistance, rolling resistance etc. then they may be grouped under a collective heading of Tractive resistance TR. The tractive resistance is measured in Newtons (Force) and will always act in a direction to oppose motion.
JR/2008

3. Motion on an Incline Energy solution
Worked examples:
1.a A motor car of mass 1.5 tonne is moving at 50 km/hr up to the base of an incline of 1:30(sine) when the engine is switched off and the car allowed to freewheel. Calculate how far the car will move up the slope before coming to rest. Ignore frictional resistance.
Soln.
Car is moving therefore the energy is kinetic (KE = ½ mv2)
When the car stops it has gained height and possess potential energy (PE = mgh)
As no energy losses are incurred it is a direct transfer KE to PE
½ mv2 = mgh transpose to find the height gain h
h = mass m is constant giving h = when v = 50 km/hr ÷ 3.6 = ms-1
h = = 9.83 metres of height gained.
distance moved up the slope s = height x 30 (slope 1:30 sine) = 9.83 x 30 = 295 metres
JR/2008

4. Motion on an Incline Energy solution
1.b As previous but against a frictional (tractive ) resistance of 300 Newtons
per tonne. Mass m = 1.5 tonne = 1500 kg
Soln.
As before initial KE transferred into PE BUT some losses occur as work is done
against TR when TR = 300 N/t = 300N/t x 1.5t = 450N
v = 50 km/hr ÷ 3.6 = ms-1 & height h = distance s/30 (slope 1:30 sine)
Energy equation becomes:
KE = PE + work done against TR
½ mv2 = mgh + (TR x distance s) substitute s/30 for h
½ mv2 = (m g s/30) + (TR x s)
½ x 1500 x = (1500 x 9.81 x s/30) + (450 x s)
J = (490.5 s) + (450 s)
J = s
Therefore distance s = ÷ = 154 metres
JR/2008

5. Motion on an Incline Energy solution
Summary:
For simple problems without losses the energy balance is KE = PE
If losses are incurred then the energy balance must include work done to overcome the tractive resistance (work = force x distance)
Tractive resistance (TR) is often given in terms of Newtons per tonne, care must be taken to convert this ratio into an actual force value.
The incline when expressed as a sine ratio shows the link between distance moved along the slope and vertical height gain.
eg. for a slope of 1:5 sine the distance moved along the slope s would be equal to 5 x the vertical height gained h ( s = 5h and h = s/5)
JR/2008

6. Motion on an incline Student examples:
2. A car of mass 1 ½ tonnes is at rest 400m up an incline of 1:30 (sine) when the brake is released and the car allowed to freewheel down the slope and onto a flat stretch of road.
Find the velocity at the base of the slope and the maximum distance moved along the flat if:
a. No tractive resistance is present (no friction, air resistance, rolling resistance etc.!!!!!)
b. Average tractive resistance amounts to 200 Newtons per tonne (200 N/t)
(16.17 ms-1, ∞, 10.1 ms-1, 255m)
3. A car of mass 1.2 tonnes moving at 72 km/hr is allowed to free wheel along a flat stretch of road against an average tractive resistance of 350 N/t. Calculate the maximum distance which may be covered by the car. (571.4m)
4. A car of mass 1 tonne is moving along the flat at 108 km/hr and is halted by a brake application in a distance of 300m. Find
a. the average brake force which must be supplied if TR is zero kN
b. the average brake force which must be supplied if TR is 500N/t 1 kN
JR/2008

7. Motion on an incline Worked examples:
5. A train of mass 150 tonnes approaches a gradient of 1:50 (sine) at 72 km/hr and the power is turned off. If the train has an average tractive resistance of 100 N/t calculate the distance moved up the incline (676m)
6. The train in Q5 is allowed to freewheel back down the incline from its resting position with the TR remaining constant. Find:-
a. velocity at the bottom of the incline ms-1
b. distance moved along the flat m
7. A car of mass 1 tonne is accelerated from rest to 108 km/hr in a distance of 225m up an incline of 1:15 (sine) against an average tractive resistance of 500 N/t.
Find
a. the average tractive effort which must be supplied by the car propulsion system.
b. the average velocity of the car
the average power supplied (3.154 kN, 15ms-1 , kW)
8. Students are to check answers/methods against examples previously worked using the Laws of Motion and compare methods for ease and efficiency
JR/2008