1. Work Done = Force x Distance
Work and Energy
Work-Energy Equivalence means . . .
work done on a body is equal to energy gained
and work done by a body equals energy lost.
Work Done = Force x Distance
We can use this to derive the formulas for Kinetic and Potential Energies.
Consider the work done on a body at rest, with mass m, to accelerate it over a smooth surface for a distance d.
W = F x d
But a = F m
v2 = u2 + 2as
s = v2 2 F m
W = F x mv2 2F
v2 = 2as
W = 1 2 mv2
s = v2 2a
s = mv2 2F
All Work Done = Ek = 1 2 mv2

2. Consider the work done on a body at rest, with mass m, to lift it through height h, without any acceleration or friction.
W = F x d
F = mg and d = h h
W = mgh mg

3. i) Work against gravity: W = F x d W = mgsinΘ x d W = 1 x g x 7 25 x 5
Example 1
What work is done moving a mass of 1kg steadily up slope of length 5m, given that the sin Θ = and μ = 5 12
(i) against gravity
(ii) against friction
i) Work against gravity:
W = F x d Θ 7 25
W = mgsinΘ x d
1kg 5m Θ
W = 1 x g x x 5
W = g x 7 5
252 −72
=24 P
W = 13.7J
Forces are in equilibrium FR
mgcosΘ
ii) Work against friction: mg
P = FR + mgsinΘ
W = μmgcosΘ x d
mgsinΘ
P = μR + mgsinΘ
W = x 1 x g x x 5
P = μmgcosΘ + mgsinΘ
W = 2g
= 19.6J
friction
gravity

4. Example 2
A 2kg mass falls vertically past point A at 1ms-1 and point B at 4ms-1 in a viscous liquid, with a constant resistance force of 9.6N.
Find the distance AB
F = mg – 9.6
v2 = u2 + 2as A B
1ms-1
4ms-1 mg
9.6N
F = 10N
42 = x 5 x s
a = F m
= 10s
a = 10 2
= 5 ms-2
s = 1.5m

5. An alternative solution uses Energy
At A:
Ek = 1 2 mv2
EP = mgh A B
1ms-1
4ms-1 mg
9.6N
Ek = x 2 x 12
EP = 19.6h
Ek = 1J
Ek = 1 2 mv2
At B:
EP = mgh
EP = 0J as h=0
Ek = x 2 x 42
Ek = 16J
h = 16 + Work done against resistance
h = h
10h = 15
h = 1.5m

6. Work Done when Force Applied in different direction to Travel.
When Force and direction are in the same direction we find the Work Done by multiplying them.
If they are in different direction however:
Force OR F Θ Θ
Direction of travel d
The amount of the Force going in the direction of travel is FcosΘ
Work Done = F . d
Work Done = FcosΘ x d

7. ( ) ) ( ) ( Work Done = F . d Work Done = . = Work Done = 4x8 + (-3)x2
Example
1) A force of (4i - 3j) N moves an object from A (1,2) to B(9,4) along a smooth surface. Distances in metres.
What is the work done by this force? ( )
9 - 1
4 - 2
Work Done = F . d
AB =
Work Done = ) ( 8 2 4 -3 ) ( 8 2 =
Work Done = 4x8 + (-3)x2
Work Done = 26J

8. a = F m = 6i + 2j 2 = 3i + j ms-2 v = u + at vB = 0 + (3i + j) x 3
Example
2) At time t=0 a 2kg mass is at rest at A(2i + 3j). A constant force (6i + 2j)N causes an acceleration. 3 seconds later it passes B.
a) acceleration
b) vB
c) Kinetic Energy at B
a = F m
= 6i + 2j 2
= 3i + j ms-2
v = u + at
vB = 0 + (3i + j) x 3
vB = 9i + 3j ms-1
Ek = 1 2 x2x( 90 )2
|vB| =
|vB| = 90
Ek = 90J

9. Example
2) At time t=0 a 2kg mass is at rest at A(2i + 3j). A constant force (6i + 2j)N causes an acceleration. 3 seconds later it passes B. d)
e) Position vector of B
s = ut at2 AB
(3i + j) x 32
AB =
27 2 i j m
AB = OA
AB =
OB -
OB =
AB + OA ) (
27 2
9 2 ) (
31 2
15 2 ) ( 2 3 +
OB = =
= 15.5i + 7.5j m

10. Example
2) At time t=0 a 2kg mass is at rest at A(2i + 3j). A constant force (6i + 2j)N causes an acceleration. 3 seconds later it passes B.
f) Work done by the force
Work Done = F . d
Work Done = )
27 2
9 2 ( 6 2
Work Done = 6x x 9 2
Work Done = 90J

11. Page 265 Ex 11E
Q1-4
Q8-24 (even only)

12. Homework: Sheet HW 6.1

18. 4

21. F = P v Vehicles in Motion
A vehicle travelling along a road has a forward force F acting on it which is produced by an engine working at a constant rate of P watts.
Power is the measure of the rate at which work is being done.
Power = Work Done per second
P = Force x distance moved per second
P = Force x speed
P = Fv
Therefore the Force produced by an engine can be given by:
F = P v

22. The maximum velocity occurs when the acceleration is zero.
Example
1) The engine of a car is working at a constant rate of 8kW. The car has a mass of 1500kg is being driven along a level straight road against a constant resistive force of 425N. Find the acceleration of the car when its speed is 10ms-1 and it’s maximum speed.
Fc = P v
v = 10ms-1
425N
The maximum velocity occurs when the acceleration is zero.
1500kg
Fc = P v
a = F m
Therefore Fc must equal 425N
Fc =
a =
vmax = P Fc
Fc = 800N
a = 0.25ms-2
vmax =
Resultant Force =
vmax = 18.8ms-1
F = 375N

23. vmax occurs when the acceleration is zero. vmax = 21000 1060
Example
2) A car travelling along a level road against a resistive force of 700N, has a mass of 1800kg and a maximum speed of 30ms-1. If the car works at the same rate and the resistive force is unchanged find the maximum speed when ascending a slope of sin
Fc = P v
v = 30ms-1
Fc = P v R
700N
1800kg
1800kg
700N
sin
P = Fvmax
P = 700x30
1800g
1800gsinΘ
P = W
vmax = P Fc
Fc = gsinΘ
vmax occurs when the acceleration is zero.
vmax =
Fc = 1060N
vmax = 19.8ms-1

24. Page 271
Ex 11G

25. 1)

27. 2)

29. 3) 4)

33. 5) 6)

36. p = m x v j = Δp j = pF -pI also j = F x t Momentum and Impulse
Momentum, often given the symbol p, is the product of the mass and velocity of an object. It has the units Newton-seconds (Ns).
In a closed system the total momentum always remains constant.
Impulse is the difference in momentum caused by a Force acting on the object(s) over a time t. This is often given the symbol j.
p = m x v
j = Δp
j = pF -pI
also j = F x t
Momentum and Impulse can be scalar or vector quantities.

37. The 2 masses collide and coalesce (stick together)
Example 1)
A 1kg mass with velocity (2i+3j) ms-1 meets a 2kg mass with v = (-2i+5j) ms-1
The 2 masses collide and coalesce (stick together)
What is their final speed and direction?
Before Collision
After Collision
mt = 3kg
ma = 1kg
mb = 2kg
vt = ? ms-1
va = 2i+3j ms-1
vb = -2i+5j ms-1
pI = ma x va + mb x vb
pf = mt x vt
pI = 1(2i+3j) + 2(-2i+5j)
pf = 3vt
pI = -2i + 13j Ns
3vt = -2i + 13j
vt = i j ms-1

38. Example 2)
A blue ball with velocity (4i+j) ms-1 hits a stationary white ball of the same mass which then moves off with velocity v = (3i+2j) ms-1.
a) What is the final velocity of the blue ball?
b) What is the angle between the two final velocities?
Before Collision
After Collision
ma = mkg
mb = mkg
ma = mkg
mb = mkg
ub = 0 ms-1
va = ? ms-1
vb = 3i+2j ms-1
ua = 4i+j ms-1
pf = ma x va + mb x vb
pI = ma x ua + mb x ub
pf = mva + m(3i+2j)
pI = m(4i+j) + 0
pI = 4mi + mj Ns
4mi + mj = mva + m(3i+2j)
4i + j = va + 3i+2j
va = i - j ms-1

39. va = i - j ms-1
cosΘ = va.vb |va||vb|
vb = 3i+2j ms-1
cosΘ = 1x3 +( −1 x2) ( 12+ −1 2 )( )
cosΘ = x 13
Θ = …
Θ = 79o

40. Force on a Surface
When a jet of water hits a wall and does not rebound the loss of momentum is equal to the impulse from the wall. If we know how much water is hitting the wall we can find the momentum destroyed.
Example:
Water comes out of a horizontal pipe and hits a wall without rebounding. The CSA of the pipe is 8cm2 and the water comes out at 6ms-1. Find the magnitude of the force.
Find volume of water
Volume in 1 sec = CSA x v
8cm2
Volume in 1 sec = x 6
v = 6ms-1
Volume in 1 sec = m3

41. Find mass of water
ρ of water is 1000kgm-3
m = V x ρ
m = x 1000
m = 4.8kg
Find momentum
p = m x v
j = F x t
p = 4.8 x 6
As p = j and t = 1sec
p = 28.8 Ns
28.8 = F x 1
F = 28.8N