M1, C2 - Work, Energy & Power

Question 1

What is the formula for work done

Answer 1

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.)

Question 2

What is the unit for work done

Answer 2

Joules
J

Question 3

What is the formula for wok done agaisnt gravity

Answer 3

Wd = mgh
where h is the vertical distance raised, g is the acceleration due to gravity, mass it the mass of the particle

Question 4

A particle moves at constant speed with constant resistance to motion R. When the particle moves 12m, the wok done is 96J, calculate R

Answer 4

Wd = F x d
F = 8N
F = R = 8N

Question 5

How do you calculate the work done against gravity

Answer 5

Vertical height moved up = h
Sub into mgh = ans

Question 6

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

Answer 6

Work done = F x d
F = (coefficient of friction) * R
Find F and multiply by parallel distance to the slope

Question 7

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

Answer 7

Work done = component of force in direction of motion x distance
Work done = 27 * cos(25) x 15 = 367J

Question 8

What is the formula for kinetic energy

Answer 8

K.E. = 0.5 * m * v^2

Question 9

What is the formula for potential energy

Answer 9

P.E. = mgh where h is the distance above an arbitrary fixed level

Question 10

What is change in kinetic energy equal to

Answer 10

Work done = change in K.E.
(0.5 * m * (v^2 - u^2))

Question 11

How can you equate change in E.K. and force x distance

Answer 11

F x d = 0.5 * m * (v^2 - u^2)

Question 12

When does the sum of the particle’s kinetic energy and potential energy remain constant

Answer 12

When no external forces (other than gravity) do work on a particle during it motion

Question 13

What is the change in total energy of a particle equal to

Answer 13

Work done on particle = change in total energy of a particle

Question 14

When there is an external force (other than gravity) on a particle, what happens to the total energy of the particle and why

Answer 14

Total energy of the particle decreases as some of the energy is doing work against the resistive force

Question 15

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

Answer 15

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

Question 16

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

Answer 16

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

Question 17

What is the equation for power

Answer 17

Power = driving force x speed
P = F * v

Question 18

What is power the rate of

Answer 18

Power is the rate of doing work

Question 19

What is the unit for power

Answer 19

Watts
W

Question 20

What is the driving force F equal to in terms of power

Answer 20

F = P / v

Question 21

What is acceleration when speed is at a maximum

Answer 21

a = 0ms^-2

Question 22

What is the resultant force when an object moves at maximum speed

Answer 22

0
F = ma
a = 0
Therefore F = 0

Question 23

What is the resistive force equal to when an object is at maximum speed

Answer 23

a = 0, therefore no resultant force
Resistive force = pulling force

Question 24

A van of mass 1250kg with engine of power 24kW has a constant resistance to motion of 600N. Calculate the maximum speed

Answer 24

At max speed acceleration = 0, thus resultant force = 0
Resistive force = pulling force = 600N
F = P / v
600 = 24000 / v
v = 40ms^-1

Question 25

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

Answer 25

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

Question 26

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

Answer 26

Gravity (weight * sin(x))
Resistance forces such as mu * R

Question 27

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

Answer 27

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