M1, C2 - Work, Energy & Power Flashcards
What is the formula for work done
Work done = component of force in direction of motion x force moved in distance of force
Wd = f x d
Wd = fcos(x) x d
(Wd = change in E.K.)
What is the unit for work done
Joules
J
What is the formula for wok done agaisnt gravity
Wd = mgh
where h is the vertical distance raised, g is the acceleration due to gravity, mass it the mass of the particle
A particle moves at constant speed with constant resistance to motion R. When the particle moves 12m, the wok done is 96J, calculate R
Wd = F x d
F = 8N
F = R = 8N
How do you calculate the work done against gravity
Vertical height moved up = h
Sub into mgh = ans
How do you calculate the work done against friction up a slope if you know the coefficient of friction and the mass of the particle
Work done = F x d
F = (coefficient of friction) * R
Find F and multiply by parallel distance to the slope
A sledge is pulled 15m across a smooth sheet of ice by a force 27N. The force is inclined at 25 degrees to the horizontal. Calculate the work done by the force
Work done = component of force in direction of motion x distance
Work done = 27 * cos(25) x 15 = 367J
What is the formula for kinetic energy
K.E. = 0.5 * m * v^2
What is the formula for potential energy
P.E. = mgh where h is the distance above an arbitrary fixed level
What is change in kinetic energy equal to
Work done = change in K.E.
(0.5 * m * (v^2 - u^2))
How can you equate change in E.K. and force x distance
F x d = 0.5 * m * (v^2 - u^2)
When does the sum of the particle’s kinetic energy and potential energy remain constant
When no external forces (other than gravity) do work on a particle during it motion
What is the change in total energy of a particle equal to
Work done on particle = change in total energy of a particle
When there is an external force (other than gravity) on a particle, what happens to the total energy of the particle and why
Total energy of the particle decreases as some of the energy is doing work against the resistive force
A particle of mass 2kg is projected with speed 8m/s up a line of greatest slope of a rough plane inclined 45 degrees to the horizontal. Coefficient of friction between the particle and the plane is 0.4. Calculate the distance travelled up the plane before the particle comes to instantaneous rest
Loss in E.K. = Gain in E.P. + Work done
0.5 * 2 * 64 = 2g * sin(45) * x + 0.4 * R * cos(45) * x
Where R is the normal reaction = 2g * cos(45) and x is the unknown distance travelled
x = 3.3m
A skier moving downhill passes point A at 6m/s. Descends 50m vertically then ascends 25m to point B at speed 4m/s. Skier of mass 55kg and travels 1400m from A to B. Non-gravitational resistances to motion of magnitude 12N. Calculate the work done by the skier
Loss of E.K. = 0.5 * 55 * (4^2 - 6^) = -550J
Loss of E.P. = 55 * 9.8 * (25 - 50) = -13475J
Work done against friction = 1400 x 12 = 16800N
Work done against friction - ‘free energy’ = work done by skier
16800 - 550 - 13475 = 2775J
What is the equation for power
Power = driving force x speed
P = F * v
What is power the rate of
Power is the rate of doing work
What is the unit for power
Watts
W
What is the driving force F equal to in terms of power
F = P / v
What is acceleration when speed is at a maximum
a = 0ms^-2
What is the resultant force when an object moves at maximum speed
0
F = ma
a = 0
Therefore F = 0
What is the resistive force equal to when an object is at maximum speed
a = 0, therefore no resultant force
Resistive force = pulling force
A van of mass 1250kg with engine of power 24kW has a constant resistance to motion of 600N. Calculate the maximum speed
At max speed acceleration = 0, thus resultant force = 0
Resistive force = pulling force = 600N
F = P / v
600 = 24000 / v
v = 40ms^-1
A car of mass 1100kg travels at constant speed 15ms^-1 along a road at incline 7 degrees to the horizontal. Engine works at a rate of 24kW.
a) Calculate the magnitude of the non-gravitational resistance to motion
b) Calculate the initial acceleration if the engine rate of work is increased to 28kW
a) Driving force = P / v = 24000 / 15 = 1600
Frictional force + force against gravity = driving force
F + 1100g * sin(7) = 1600
F = 286N
b) F = ma
F = (28000 / 4) - 286 - 1100g * sin(7) = 1100 * a
a = 0.242ms^-2
When travelling up an inclined plane, what 2 resistances work against a particle that a driving force would equal the sum of if the speed is constant
Gravity (weight * sin(x))
Resistance forces such as mu * R
A car has mass 2600kg and at speed v ms^-1 the total resistances to motion have magnitude (800 + 5v^2)N. Cruise control means engine maintains constant speed of 18ms^-1. Find power generated by engine when car travels at 4 degree incline to the horizontal
v = 18
Resistance forces = (800 + 5(18)^2) = 2420N
Gravitational resistance = 2600g * sin(4) = 1777.4
Total driving force = 2420 + 1777.4 = 4197.4N
P = F * v = 4197.4 * 18
P = 75553W
