Document From Marzia
Document From Marzia
Document From Marzia
Prepared By:
Ahmed Saya
Answers available at
www.eastpk.org/cars
KINEMATICS
Kinematics means motion of particles with respect to time.
Scalar quantities: {Distance, Speed and time} only magnitude is important irrespective of the
direction.
10
7
Total = 10 + 7 = 17 units
Vectors quantities: {Displacement, velocity, acceleration} magnitude and direction both are
important.
10
3
7
Vector = 3 units
Distance: It is the length of the movement of particles since its start irrespective of the direction.
Speed: It is the rate of change of distance i.e. the rate at which distance is covered.
Displacement: It is the shortest distance from the starting point.
Velocity: It is the rate of change of displacement but it can be both positive and negative.
Acceleration: It is the rate of change of both, velocity as well as speed.
(hrs. or sec.)
S=D
T
(Kmh-1 or ms-1)
To convert from:
Kmh-1 to ms-1
X 1000
3600
To convert from:
Ms-1 to kmh-1
X 3600
1000
Average speed =
Total distance
Total time
Example 1:
A cyclist is travelling at a speed of 20 ms-1 for 50 seconds on a dirt track. He then travels for 1500 meters on
a proper track at 50 ms-1. Finally he travels for 900 meters for 30 seconds on a wet track. Find:
i) the distance travelled on the dirt track
ii) the time taken on the proper track
iii) the speed on the wet track
iv) average speed of the whole journey
Solution:
i) distance = 20 X 50 = 1000 meters
ii) time = 1500 / 50 = 30 seconds
iii) speed = 900 / 30 = 30 ms-1
iv) Average Speed = 1000 + 1500 + 900
50 + 30 + 30
3400
110
= 30.91 ms-1
Distance time
Distance (m)
75
25
A
25
10
30
Speed (m s-1)
50
A
10
50
90
Time (seconds)
Displacement (m)
50
A
80
10
Time (seconds)
100
120
Velocity (m s-1)
x
A
D
Time (s)
y
E
EXAMPLE 2
A car is travelling at a constant speed of 72 km h-1 and passes a stationary police car. The police car
immediately gives chase, accelerating uniformly to reach a speed of 90 km h-1 in 10s and continues at this
speed until he overtakes the other car. Find
i)
ii)
The v t graphs of the car and the police are shown in the following figure
V m s-1
25
A
car
20
Police car
C
10
t sec
ii)
Starting point
0
Turning point, v = 0. Rest
Gone
movement in
Beyond
opposite direction
The Starting
Point
If X > Y, the particle has not yet reached the starting point, displacement is positive.
X
Y
If X = Y, the particle is at the starting point, displacement is zero.
X
Y
If X < Y, the particle has gone beyond the starting point, displacement is negative.
X
Velocity (m s-1)
(s)
u
t
(o, u) (t, v)
v u = acc
to
vu=a
t
v u = at
v = u + at
x[u+v]xt=S
S=[u+v]t
S=(u+v)t
S = (u + u + at) t
S = x ( 2ut ) + at2
S = ut + at2
v = u + at
v u = at
T = v u
a
S=(u+v)t
S=(v+u)(vu)
a
S = v2 u2
2a
v2 u2 = 2as
Time (s)
EXAMPLE 3
A particle is travelling at a constant acceleration of 0.06 ms-2. His initial velocity is 3 ms-1 and after t seconds
has a velocity of 6 ms-1. Find
(a) the total time taken for the journey.
(b) the total distance travelled
A to B: constant a = 0.06 ms2
a)
a = v2 v1
T.T
T . T = 6 3 = 50s
0.06
b)
Distance = x 50 x (3 + 6) = 225 m.
EXAMPLE 4
Two particles P and Q are moving on the same horizontal line towards each other. P passes a point on the
line with speed 8 m s-1 and constant acceleration 0.75 m s -2. Simultaneously, Q passes a point B on the line
with speed 4 m s-1 and constant acceleration 1.25 m s -2. Given that distance AB is 64 m ,calculate
i)
ii)
Solution:
i)
ii)
4.
Gravity:
Gravity is the acceleration with magnitude 10ms-2
Dropped:
Whenever the particle is dropped, it will always have an initial speed of 0. When the particle is
dropped it will fall vertically downwards.
Thrown downwards:
When the particle is thrown downwards, it will definitely have some initial speed.
Thrown upwards:
When the particle is thrown upwards, it will always move vertically upwards. It must have some
initial speed.
When particle is moving downwards, its gradient is positive i-e. 10 m/s.
When particle is moving upwards, its gradient is negative i-e. -10 m/s.
When a particle falls freely under gravity. Its motion is symmetrical. If the motion is not obstructed
by any hurdle.
Max point
v=0
S, t, g = -10
u=u
s, t, g = 10, v = v
When the motion of the particle is obstructed then the particle will never reach the maximum
height, instead it will reach the greatest height.
Max height
Greatest height
EXAMPLE 5
A particle is projected vertically upwards with a velocity of 30 m s-1 from a point O. Find
a) The maximum height reached
b) The time taken for it to return to O
c) The time taken for it to be 35 m below O
Solution:
Take positive direction as vertically upwards; the initially velocity u is 30 m s -1
a) At max height, v = 0; acceleration is -10 m s -2.
Using equation v2 = u2 + 2as
We get h = 45 m
hence max height is 45 m
b) When particle returns to O, displacement is 0
Using equation s = ut + at2
We get 5t2 + 30t = 0; t = 0 (at start) and t = 6 (when particle returns to O)
c) When particle is 35 m below, displacement is -35 m
Using equation s = ut + at2
We get t = -1 (not admissible) and t = 7
Hence particle will take 7 seconds
5
Motion of the particle with varying acceleration.
If acceleration is varying with respect to time then velocity and displacement will also vary with
respect to time and hence in order to solve such questions, we have to apply the Calculus i.e.
differentiation and integration.
Displacement
v=s
ds = v
dt
Velocity
a=v
dv = a
dt
Acceleration
14. When you are told to find the distance, follow the following steps: Step 1: Check whether the particle is turning within the required time.
Step 2: If the particle does not possess the turning or does not turn within the
required time, the distance is equal to displacement.
Step 3: If the particle turns within the required time; the distance is not equal to
displacement and hence to find distance, the following steps are to be followed.
(a)
(b)
displacement.
(c)
If b > 0
a
b
a-b
Distance = a + a b = 2a b
(d)
If b = 0
a
a
Distance = 2a
(e)
If b < 0
a
b
a
Distance = 2a + b
15. When the distance from 0 is required, it means that the question is asking for displacement.
Example 6
A particle travels in a straight line from a point A to a point B. its velocity t seconds after leaving A is v m s-1,
where
V = 4t 0.04t3
Given that AB is 100 m, find
a) the value of t when the p
article reaches B
b) whether the particle is speeding up or slowing down at the instant that it reaches B
N = 4t 0.04t3
S = 4t 0.04t3
S = 4t2 0.04t4
2
4
2
s = 2t 0.01t4
When s = 100
2t2 0.01t4 = 100
0.01t4 2t2 + 100 = 0
0.01t2 2t + 100 = 0
T = 100
t2 = T
ii)
In order to determine whether the particle is speeding
t2 = 100
up or slowing down check the sign of acceleration, if acc. is positive,
t = 10sec
the particle is speeding up and if acc. is negative particle is slowing
3
ii)
v = 4t 0.04t
down.
a = 4 0.12t2
when t = 10
a = 4 0.12(10)2
=-8
As a is negative, the particle is slowing down.
Note: In order to determine whether the particle is speeding up or slowing down check the
sign of acceleration, if acc. is positive, the particle is speeding up and if acc. is negative particle is
slowing down.
Exercise 1
1. A train moves with a constant speed of 20 m s-1 for 10s. it then accelerate uniformly for next 5s to
reach a speed of V m s-1 when it decelerates uniformly to rest in the next 4s.
a) Sketch a v t graph for the journey.
b) If the total distance covered is 385m, find the value of V.
2. Two particles A and B are moving in the same direction on parallel horizontal lines. They pass a
certain point O at the same time, A moving with the speed of 3 m s-1 and constant acceleration 0.5
m s-2 and B moving with the speed of 4 m s-1 and constant acceleration 0.2 m s-2+.
If the speed of A and B are equal after t seconds, calculate
a) The value of t
b) The distance between the particles at that instant.
3. A stone is dropped from the top of a building of height 20 m. find the time it takes to reach the
ground and the velocity with which it hits the ground.
4. From the top of a vertical tower 246m high, a stone is projected vertically downwards with a speed
of 8 m s-1. After t seconds another stone is projected vertically upwards from the level of the base of
the tower with a speed of 25 m s-1. Given that two stones are first at the same height 6 seconds
after the projection of the first stone, calculate
a) The value of t
b) The velocity of the second stone at this instant
5. A particle is projected vertically upwards from a ground with a speed of 36 m s-1. Calculate
a) The time for which it is 63 m above the ground
b) Speed which it has at this height on its way down
c) Total time of flight
6. A particle is travelling in a straight line with constant acceleration a ms-2. It passes a point A with
speed u ms-1 . between 3 and 4 seconds after passing A it travels 11.4m and between 4 and 5
seconds after passing A it travels 11.8 m. find the value of u and of a.
VECTORS
Parallelogram Rule.
Example1
Example 2
5km
5km
5.44 km
5.44m
21.80
21.80
2 km
Example 3
Find the resultant of the following vectors?
a) 10m 0600. Followed by 4m east.
b) 40km on the bearing of 600 and 40 km on the bearing of 1100.
a)
120 90
10m
60
13.6m
150
8.5
4m
Exercise 2
1. Two forces, each of magnitude 10 N , act at a point O in the directions of OA and OB, as shown in
the diagram. The angle between the forces is . The resultant of thes two forces has magnitude 12
N.
a) Find
b) Find the component of the resultant force in the direction of OA
10 N
10 N
2. Two horizontal forces, of magnitudes 10 N and P N, act on a particle. The forces of magnitude 10 N
acts due west and the forces of magnitude of P N acts on a bearing 030 (see diagram). The resultant
of these two forces acts due north. Find the magnitude of this resultant.
PN
30
10 N
FORCES IN EQUILIBRIUM
What is force?
Force is a push or pull exerted on a particle in order to move the particle or stop the particle.
Types of Forces:
1. Weight:
It is the gravitational attraction of the earth on the particle. Weight always act
vertically downwards ( ) and the formula to obtain weight is:
w=mxg
N = kg x ms-2
e. g. 2kg
w = 2 x 10 = 20
3. Air Resistance:
If the particle is not on the surface then instead of normal reaction air resistance
exists. If the particle is falling down, air resistance will act vertically upwards and if the
particle is going up, air resistance will act vertically downwards.
Air resistance
Air Resistance
4. Tension:
When the particle is being pushed or pulled through a string, then the force exerted
on the string is called tension.
Tension
f f.
f f.
What is Equilibrium?
Condition of equilibrium occurs when the particle is either travelling at a constant speed or
is at rest. In both of these conditions, acceleration is equal to zero. Hence equilibrium occurs when
acceleration = 0.
5N
XN
s+x=8
x = 3N
8N
Y
12
2Y
y + 12 = 2y
y = 12N
2x
8 + 3y
5y
X 2 8
5y = 8 + 3y
2x = x + 2 + 8
3- If the force applied to the particle is an inclined force then first we will resolve the
force and then equate the upward forces with the downward forces.
F sin
F cos
EXAMPLE 1
NR
F
5kg
F sin 30
ff = 0
30
Fcos30
300
300
50 cos 30
50 sin 30
50N
Smooth Surface
=0
ff = 0
Fcos30 = 50sin30
F = 50sin30
Cos30
= 28.9N
R = Fsin30 + 50cos30
= 28.9sin30 + 50cos30
= 57.8N
Note:
Angle between the weight component and resolving component will always be equal to the angle of
slope.
EXAMPLE 2:
A particle of mass 1kg rests on a horizontal floor. The coefficient of friction between the particle and the
floor is 0.5. What force is required just to make the particle move when:
a) Pulling horizontally.
b) Pulling at an angle of 300 to the horizontal.
Note:
When you are encountered with the following terms, it is an indication that the particle is in the
condition of equilibrium.
i)
Particle is at rest.
ii)
Particle is travelling at a constant speed.
iii)
Particle is at maximum speed.
iv)
Particle is at momentary rest.
v)
Particle is about to move.
vi)
Particle is about to slip.
vii)
Particle is travelling at a terminal velocity.
R
solution a)
ff = 0.5R
1kg
10N
R = 10N
0.5R = x
0.5(10) = x
x = 5N
solution b)
F
Fsin30
300
Fcos30
ff = 0.5R
10N
R + Fsin30 = 10
R = 10 1 F
2
ff = 0.5(10 1 F)
2
Fcos30 = 5 1 F
4
F = 4.48N
R = 10 0.5(4.48) = 7.76N
LAMIS THEOREM:
o
Note:
LAMIs theorem is an alternative method of resolving forces when the particle is in
equilibrium. The formula is:
Sin A =
Sin B =
Sin C
a
b
c
EXAMPLE 3
A particle of weight 10 is suspended by 2 strings. If these strings ma e angles of 30 and 40 to the
horizontal. Find the tensions in the string.
Solution:
Let the tension in the string be T N and P N respectively.
Using Lamis theorem,
T
10
=
=
sin 0 40 ) sin 0 30 )
sin 110
Giving T = 8.2 N and P =9.2 N
P N
T N
40
30
10 N
Where a, b and c are three forces and A, B and C are angles between the forces.
Diverge
Converge
The resistive force is greater than the driving force; the value of resistive force will be equals
to the driving force.
F.F = 10
F.F = 10
D.F = 10
D.F = 15
7
f.f = 10
D.F = 7
Exercise 3
1. particle of mass 1kg is placed on a rough plane inclined at an angle of 300 to the horizontal. The
coefficient of friction is 2/5. Find the least force parallel to the plane that is required
a) To hold the particle at rest.
b) To make the particle slide up the plane.
2. Find sum of the components of the forces acting as shown in the figure in the direction of the dotted
lines and hence find their resultant
20 N
10 N
30
0
30
30 N
3. Two forces P N and Q N include an angle of 120 and their resultant is 1 . if the included angle
between the forces is 0 , their resultant would be
. find and
4. A box of mass 1 g is placed on an inclined plane of angle 0 and is ust held there at rest by a
horizontal force P. if the coefficient of friction is 0.4, find P.
5. A hori ontal force of 10 ust prevents a mass of 2 g from sliding down a rough plane inclined at
4 to the hori ontal. Find the coefficient of friction.
6. Figure shows a small block of mass 5 kg held against a rough vertical wall by a force
inclined at
and angle of 0 to the wall. If the coefficient of friction is 0.4 , calculate the value of when the
block is
a) Just prevented from slipping down
b) Just about to move up the wall
PN
7. The three coplanar forces shown in the following figure are in equilibrium. Calculate the value of
and P
3P N
30 N
120
2P N
8. Two rectangular boxes A and B are of identical size. The boxes are at rest on a rough horizontal floor
with A on top of B. Box A has mass 200 kgand box B has mass 250 kg. A horizontal force of
magnitude P N is applied to B (see diagram). The boxes remain at rest if 31 0 and start to move
if P > 3150.
a) Find the coefficient of friction between B and the floor.
The coefficient of friction between two forces is 0.2. Given that P > 3150 and that no sliding takes
place between the boxes .
b) Show that the acceleration of the boxes is not greater than 2 m s-2
c) Find the maximum possible value of P.
3 LAW OF MOTION:
To every action there is an equal and opposite reaction.
EXAMPLE 1
A horizontal force of 0.5N acts on a particle of mass 0.2kg and the frictional force is 0.2N. Find the
acceleration of the particle.
0.5 0.2 = (0.2) a
0.2
0.5N
0.3 = 0.2a
a = 0.3 = 1.5ms-2
2N
0.2
EXAMPLE 2
A force of 1N acts on a particle of mass 2kg which is initially at rest. Find the resulting acceleration and the
velocity of the particle after 5 seconds.
1
F = ma
1 = (2) a
a = 0.5ms-2
20
v = u + at
v = 0 + 0.5(5)
v = 2.5ms-1
D.F R.F = ma
5kg
50 T = 5a
5kg
T 50 = 5a
EXAMPLE 3
A horizontal force of 0.5 N acts on a body of mass 0.2 kg. there is a frictional force of 0.2 Nopposing the first
force, what acceleration will be produced?
We first find the resultant or the net of the forces acting: 0.5 0.2 = 0.3
F = ma
0.3 = 0.2 a ; giving a = 1.5 m s-2
If multiple forces are applied on a particle which is not in equilibrium then we will resolve the
vertical forces and the horizontal forces and then find the resultant of the vertical and horizontal forces.
(Hook)
x
Slack
T
3kg
7kg
tout
Example 4
70 T = 7a
T 30 = 3a
70 T = 7a
-30 + T = 3a
40 = 10a
30
70
a = 4ms-2
T = 3(4) + 30 = 42N.
Tension exerted on the pulley will always be 2 times the tension:
x= 2(T) = 2 x 42 = 84N
EXAMPLE 5
The following figure shows two particles: A of mass 1 kg and B of mass 2 kg connected by a light inextensible
string which passes over a smooth pulley at C. the system is held at rest with B hanging freely while A is on a
rough plane inclined at to the hori ontal where tan = . The coefficient fo friction between A and the
plane is 0.2. find the magnitude of the acceleration of each particle and the tension in the string when the
system is released.
C
1 kg
2 kg
Solution:
Let the acceleration of B be a ms-2 downwards and hence A will move towards C with the same acceleration.
The string transmits a tension T. The figure below shows the forces acting on A and B.
C
For A, since acceleration is along AC, and the tension is opposed by the frictional force , F and the component of
weight downslope, g sin , we have
T (F + g sin ) = 1a
resolving forces perpendicular to the plane , the normal reaction is
R = g cos
the frictional force F is given by
F=R
= g cos
Substituting F into 2nd equation:
T ( g cos g sin ) = a
4
3
2(10) [(0.2)(10) 5 + (10) 5 ] = 3a
a = 4.13 ; and T = 11.74
Hence the acceleration of each particle is 4.13 m s-2 and the tension in the string is 11.74 N.
Exercise 4
1. A particle of mass 2.5 kg is moving at a steady speed of 12 m s-1 when it meets with a fixed
resistance of 10 N. How long does it take to come to rest?
2. A body of mass 4 kg is pulled from rest to a speed of 4.5 m s-1 in a time of 3 seconds on a rough
hori ontal surface by a force of 20 which ma es an angle of 10 with the horizontal. Find the
coefficient of friction.
3. PQR is a fixed wedge on level ground (as shown in the figure) where PQ = 5 m, QR = 3 m and PRQ is
a right angle. Particle A of mass 1.5 kg, lies at the foot of the smooth slope PQ, attached by a light
inextensible string passing over the smooth pulley at Q, to the particle B of mass 1 kg . B is released
from rest when it is 2 m above ground level. Find
a) The acceleration of the particles,
b) How far A will travel up the slope before coming to momentary rest.
4. A wedge has two equally rough faces each inclined at 30 to the hori ontal. Masses of g and 2 g,
one on each face, are connected by a light inextensible string passing over a smooth pulley at the
top of the wedge. The coefficient of friction between each mass and the surface of the wedge is 0.2.
Find the acceleration of the masses when they are released.
B
1.5kg
5m
2m
3m
5. The following figure shows two masses of 5 kg and 7 kg respectively connected by a light
inextensible string which passes over a smooth fixed pulley. The system is released from rest and
the 7 kg mass reaches the ground after 3 second s. Calculate
a) The acceleration of masses while the string remains taut
b) The total distance moved by the 5 kg mass before it comes instantaneously to rest,
assuming that it does not reach the pulley.
7 kg
5 kg
6. A block A of mass 3 kg is attached to a string S1. Another block B of mass 2 kg is attached to the othe
other end of the string S1, and is also attached to another light inextensible string S2. The other end
of S2 is attached to a fixed point O and the blocks hang in equilibrium below O as shown in the figure
a) Find the tension in S1 and tension in S2.
The string S2 breaks and the particles fall. The air resistance on A is 1.6 N and on B is 4 N
b) Find the acceleration of the particles and the tension in S1.
S
Example1:
A force of 60N inclined at 600 to the horizontal caused the particle to move 7m horizontally.
Find the work done by the particle.
60N
60
60cos60
W.D = 60cos60 x 7
= 210 J.
If an inclined force is applied to the particle causing it to move vertically upwards then the
word done will be:
W.D = Fsin x S
F
Fsin
Example2:
A force of 60N inclined at an angle of 300 caused the particle to move 12m vertically
upwards. Find the W.D.
W.D = 60sin30 x 12
= 360J.
If an inclined force is applied to the particle causing the particle to move in the direction of
the force then:
W.D = F x S.
S
F
POWER
Power is the rate of doing the work. It is the amount of work done within a certain time.
Hence power is calculated as:
Power = W.D = F x S = F x v
T
T
P = W.D = J
T
S
P = Watt
Hence Power can be defined as the velocity with which the particle when the force is applied to the particle.
It is the product of t and v in the direction of the movement of the particle.
ENERGY
1. Kinetic Energy.
2. Potential Energy.
3. Mechanical Energy.
1. Kinetic Energy:
When force acting on a particle of mass m (kg) gives it a velocity of v (ms-1) from
rest, it is said to be work done by kinetic energy. The kinetic energy of a particle can
thus be regarded as work done on the particle by a force in giving it a velocity from
rest. Since kinetic energy is work done, its unit is Joules.
K.E = mvs
Where m is mass and v is velocity.
If the speed changes from u ms-1 to v ms-1 then work done will be the difference
between final K.E and initial K.E.
W.D by K.E = mv2 mu2
m (v2 u2)
EXAMPLE 3
A force of 50N is being applied on a particle of 10kg giving it a speed of 15 ms-1 from rest. Find the work
done by the kinetic energy.
K.E = x 10 x 152 x 10 x 02
= 1125 J.
EXAMPLE 4
A force acting on a particle of mass 2 kg moves it from rest to a velocity of 3ms-1 over a distance of 5m. What
is the magnitude of the force applied?
M = 2kg
v = 3 u = 0.
2
`
W.D = x 2 x 3 x 2 x 02
= 9J.
W.D = F x S
9=Fx5
F = 1.8N
NOTE:
When a particle travels through steady speed it does not have a kinetic energy because there is no
change in kinetic energy.
EXAMPLE 3
A car of mass 400kg travelling at 9ms-1 comes to rest in 200m. What is the resistance?
W.D = x 400 x (02 92)
= -16200 J
NOTE:
When there is loss in kinetic energy it is an indication that some resistive force is applied to the
particle in order to stop it.
F x S = 16200
F x 200 = 16200
F = 81N
2. Potential Energy:
Potential energy is the ability to do the work because of the position of the particle
when the particle is on the ground, it has no potential energy. As particle moves up,
there is a gain in potential energy. Potential energy is also referred to as work done
against gravity and the formula to calculate P.E. is:
P.E = mgh
Where m is mass, g is gravity and h is height.
In order to find the work done by the potential energy we have to find the difference between final
potential energy and initial potential energy.
W.D by P.E = Final P.E - Initial P.E = mgH mgh
mg (H-h)
EXAMPLE 5
Find the work done when a load of mass 50kg is lifted vertically through 10m.
W.D by P.E = 50 x 10
= 5000kJ
EXAMPLE 6
A block is pulled through 4m at a steady speed by a force of 20N. The plane is inclined at an angle of 600.
Find
i) W.D by K.E
ii) W.D by P.E
4m
20N
h
600
i)
ii)
No K.E
H = 4sin60 = 2 3m
W.D by the particle = 4 x 20 = 80J
As there is zero K.E, all W.D will be by the P.E
P.E = 80J
Mgh = 80
M 80 10 x 2 3) = 2.3 g.
NOTE:
When a particle travels horizontally there is no change in potential energy and hence work done
by potential energy is zero.
3. Mechanical Energy:
Mechanical energy is the sum of kinetic energy and potential energy.
M.E = P.E. + K.E.
Mechanical energy throughout the motion of the particle is constant IF no external force is applied to it.
External force = Driving Force or Resistive force or Both driving and resistive force
When NO external force is applied, Law of Conservation of Energy is being applied to the particle. Law of
Conservation of Energy states that since mechanical energy does not change therefore change in kinetic
energy must correspond with change in potential energy, thus;
Gain in K.E = Loss in P.E
Loss in K.E = Gain in P.E
When an external force is applied to the particle law of conservation of energy does not prevail and hence
we will apply the following formula to find any possible unknown.
Work done by = Change in Kinetic Energy + change in potential energy + work done by resistive force
Example 7
A
B
The diagram shows a vertical cross section of a surface. A and B are two points on the cross section. A mass
of 0.15 kg is released from rest at A.
a) Assuming that the particle reaches B with a speed of 8 m s-1 and that there are no resistances to
motion. Find the height of A above B
b) Assuming instead that the particle reaches B with a speed of 6 m s-1 and that the height of A above B
is 4 m. find the work done against the resistance to motion.
Solution
a)
x 0.15 x 82 = 0.15 x 10 H.
H = 3.2m.
b)
W.D by D.F = in K.E in P.E + W.D by R.F
0 = (1/2 x 0.15 x 82) + (-0.15 x 10 x 4)
m = 3.3J
Example 8
The top of an inclined plane is at a height of 0.7m above the bottom. A block of mass 0.2kg is released from
rest at the top of the plane and slides a distance of 2.5 m to the bottom. Find the kinetic energy of the block
when it reaches the bottom of the plane in each of the following cases:
a) The plane is smooth
b) The coefficient of friction between the plane and the block is 0.15
R
0.7
2.5m
2.4
i)
ii)
K.E = mgh
= 0.2 x 10 x 0.7 = 1.4J.
R = 2cos
= 2(2.4/2.5) = 1.92
ff = R
= 0.15 x 1.92 = 0.288.
W.D b D.F = in K.E in P.E + W.D by Rf.
0 = x + (0.2 x 10 x -0.7) + (0.288 x 2.5)
0 = x 1.4 + 0.72
X = 0.68J
Exercise 5
1. A force acting on a body of mass 2 kg moves it from rest to velocity of 3 m s -1 over a distnace of 5 m.
what is the magnitude of the force?
2. A car of mass 1500 kg arrives at the foot of a straight hill travelling at 30 m s -1 . it reaches
the top of the hill 40 sec later at 10 3 m s-1. The length of the hill is 1000 m, and gain in the
height is 120 m. the average resistance to the motion is 500 N. Find the work done by the
engine and the average power developed.
3. A particle of mass 0. g is pro ected up an incline of angle where sin = 0. with speed 4 ms -1 .
how far will it travel up the incline if
a) The surface is smooth
b) The coefficient of friction is ?
4. The power developed by a motorcycle as it travels on a horizontal straight road at s constant speed
of 50 m s -1, is 25 kW. Calculate the resistance of the motion of the motorcycle.
5. A small block is pulled along a rough horizontal floor at a constant speed of 1.5 m s -1 by a constant
force of magnitude 30 acting at an angle of upwards from the hori ontal. Given that the work
done by the force in 20s is 20J. calculate the value of .
6.
L
N
2.45m
1.2m
The diagram shows a vertical cross section of LMN of a fixed smooth surface. M is the lowest point of the
cross section. L is 2.45 m above the level of M and N is 1.2 m above the level of M. a particle of mass 0.5 kg
is released from rest at L and moves on the surface until it leaves it at N. Find
a) The greatest speed of the particle
b) The kinetic energy of the particle at N.
The particle is now projected from N with a speed v ms -1 , along the surface towards M.
c) Find the least value of v for which the particle will reach L.
7. A lorry of mass 16 000 kg climbs from the bottom to the top of a straight hill of length 1000 m at a
constant speed of 10 m s-1 . the top of the hill is 20 m above the level of the bottom of the hill. The
driving force of the lorry is constant and is 5000 N. Find
a) The gain in gravitational potential energy of the lorry
b) Work done by the driving force
c) Work done against the force resisting the motion of lorry
On reaching the top of the hill, the lorry continues along a straight horizontal road against a
constant resistance of 1500 N . the driving force of the lorry is not now constant, and the speed of
the lorry increases from 10 m s-1 at the top of the hill to 25 m s -1 at the point P. the distance from P
to the top is 2000 m .
d) Find the driving force of the lorry while it travels to P from the top.
8. i)
A car C of mass 1200 kg climbs a hill of length 500m at a constant speed. The hill is inlcined
at an angle of to the hori ontal. The driving force exerted by Cs engine has magnitude 1800 .
find the work done against the resistance of the motionof C, as it climbs from the bottomof the hill
to the top.
ii)
Another car D, also of mass 1200 kg, climbs the same hill with increasing speed. The speed
at the bottom is 8 ms-1 and the speed at the top is 20 ms-1. Assuming the resistance to the motion of
D is constant and has magnitude 00 , find the wor done by Ds engine as D climbs from the
bottom of the hill to the top.
iii)
The driving force exerted by Ds engine is 4 times as great when D is at the top f the hill as it
is when D is at the bottom. Find the ratio of the power developed by Ds engine at the top of the hill
to the power developed at the bottom.
9. i)
A lorry P of mass 15000 kg climbs a straight hill of length 800m at a steady speed. The hill is
inclined at 2 to the hori ontal. For s ourney from bottom of the hill to the top, find
a)
the gain in gravitational potential energy
b)
the work done by the driving force
c)
the work done against the force resisting the motion
ii)
A second lorry, Q also has mass 15000 kg and climbs the hill as P. the motion of Q is subject
to a constant resisting force of magnitude, 00 , and s speed falls from 20 ms-1 at the bottom of
the hill to 10 ms-1 at the top. Find the work done by the driving force as Q climbs from the bottom of
the hill to the top.
B
5.2 m
2.5 m
10. The diagram shows a vertical cross section of ABCD of a surface. The parts AB and CD are straight
and have lengths 2.5 m and 5.2m respectively. AD is hori ontal and AB is inclined at 0 to the
horizontal. The points B and C are at the same heights abpve AD. The parts of the surface containing
AB and BC are smooth. A particle P is given a velocity of 8 ms-1 at A, in the direction AB and it
subsequently reaches D. the particle does not lose contact with the surface during the motion.
i)
Find the speed of P at B
ii)
Show that the maximum height of the cross section above AD is less than 3.2 m
iii)
State briefly why s speed is the same as that at B
iv)
The frictional force acting on the particle as it travels from C to D is 1.4N. given that the
mass of P is 0.4kg, find the speed with which P reaches D