Gravitational Field Gravitational Field Strength Gravitational Field · PDF file ·...
Transcript of Gravitational Field Gravitational Field Strength Gravitational Field · PDF file ·...
Physics Department SMK Sultan Ismail Johor Bahru
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Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational
attraction
Gravitational Field Strength The gravitational field strength at a point in the gravitational field is the gravitational force
acting on a mass of 1 kg placed at that point.
Unit: N/kg
Symbol: g
Gravitational Field Strength Formula
Gravitational Acceleration The gravitational acceleration is the acceleration of an object due to the pull of the
gravitational force.
Unit: ms-2
Symbol: g
Important notes:
Gravitational acceleration does not depend on the mass of the moving object.
The magnitude of gravitational acceleration is taken to be 10ms-2.
Gravitational Field Strength vs. Gravitational Acceleration Both the gravitational field strength and gravitational acceleration have the symbol, g
and the same value (10ms-2) on the surface of the earth.
When considering a body falling freely, the g is the gravitational acceleration.
When considering objects at rest, g is the Earth’s gravitational field strength acting on
it.
Weight The weight of an object is defined as the gravitational force acting on the object.
Unit: Newton (N)
Physics Department SMK Sultan Ismail Johor Bahru
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Differences between Weight and Mass
Weight Mass
Depends on the
gravitational field strength
Independent from the
gravitational field strength
Vector quantity Scalar Quantity
Unit Newton (N) Unit: Kilogram (kg)
Free Falling Free falling is a motion under force of gravity as the only force acting on the moving object.
Practically, free falling can only take place in vacuum.
Falling from high place
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Acceleration = 10ms-2
Initial velocity = 0
Displacement = high of the location
Launching object upward
Acceleration = -10ms-2
Velocity at maximum height = 0
Vector and Scalar Quantity A scalar quantity is a quantity which can be fully described by magnitude only.
A vector quantity is a quantity which is fully described by both magnitude and direction.
Vector Diagram
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The arrow shows the direction of the vector.
The length representing the magnitude of the vector.
Equal Vector Two vectors A and B may be defined to be equal if they have the same magnitude and point
in the same direction.
Vector Addition - Triangle Method
Join the tail of the 2nd vector to the head of the 1st vector. Normally the resultant vector is
marked with double arrow.
Vector Addition - Parallelogram Method
Join the tail of the 2nd vector to the tail of the 1st vector. Normally the resultant vector is
marked with double arrow.
Addition of 2 Perpendicular Vectors
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If 2 vectors (a & b) are perpendicular to each others, the magnitude and direction of the
resultant vector can be determined by the following equation.
Example 1
Two forces, P and Q of magnitude 10N and 12N are perpendicular to each others. What is the
magnitude of the resultant force if P and Q are acting on an object?
Answer
Use Theorem Pythagoras
F2 = 12
2 + 10
2
F = 100144
F = 15.6 N
Example 2
12 N
F 10 N
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Diagram above shows that four forces of magnitude 2N, 4N, 5N and 8N are acting on point
O. All the forces are perpendicular to each others. What is the magnitude of the resulatant
force that acts on point O?
Answer The resultant force of the horizntal component = 5 - 2 = 3N to the right
The resultant force of the vertical component = 8 - 4 = 4N acting downward.
Therefore, the magtitude of these 2 force components,
Vector Resolution
A vector can be resolve into 2 components which is perpendicular to each others.
Example 3
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Diagram above shows a lorry pulling a log with an iron cable. If the tension of the cable is
3000N and the friction between the log and the ground is 500N, find the horizontal force that
acting on the log.
Answer Horizontal component of the tension = 3000 cos30o =2598N
Friction = 500N
Resultant horizontal force = 2598N - 500N =2098N
Example 4
Diagram above shows two forces of magnitude 25N are acting on an object of mass 2kg. Find
the acceleration of object P, in ms-2.
Answer Horizontal component of the forces = 25cos45o + 25cos45o = 35.36N
Vertical component of the forces = 25sin45o - 25sin45o = 0N
The acceleration of the object can be determined by the equation
F = ma
(35.36) = (2)a
a = 17.68 ms-2
Inclined Plane
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Weight component along the plane = Wsinθ.
Weight component perpendicular to the plane = Wcosθ.
Example 5
A block of mass 2 kg is pulling along a plane by a 20N force as shown in diagram above.
Given that the fiction between block and the plane is 2N, find the magnitude of the resultant
force parallel to the plane.
Answer
First of all, let's examine all the forces or component of forces acting along the plane.
The force pulling the block, F = 20N
The frictional force Ffric = 2N
The weight component along the plane = 20sin30o = 10N
The resultant force along the plane = 20 - 2 - 10 = 8N
Vectors in Equilibrium
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When 3 vectors are in equilibrium, the resultant vector = 0. After joining all the vectors tail to
head, the head of the last vector will join to the tail of the first vector.
Forces in equilibrium Forces are in equilibrium means the resultant force in all directions are zero.
Example 6
Diagram above shows a load of mass 500g is hung on a string C, which is tied to 2 other
strings A and B. Find the tension of string A.
Answer
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Tension of string C, TC = weight of the load = 5N
All forces in the system are in equilibrium, hence
Vertical component of tension A (TA) = TC
TAcos60o = TC
TA = TC/cos60o
TA = 5/cos60o = 10N
Work Work done by a constant force is given by the product of the force and the distance moved in
the direction of the force.
Unit: Nm or Joule (J)
Work is a scalar quantity.
Formula of work
Example 1
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A force of 50 N acts on the block at the angle shown in the diagram. The block moves a
horizontal distance of 3.0 m. Calculate the work being done by the force.
Answer Work done,
W = F × s × cos θ
W = 50 × 3.0 × cos30o = 129.9J
Formula of work 2 When the direction of force and motion are same, θ = 0o, therefore cosθ = 1
Work done,
W = F × s
Example 2
Diagram above shows a 10N force is pulling a metal. The friction between the block and the
floor is 5N. If the distance travelled by the metal block is 2m, find
a. the work done by the pulling force
b. the work done by the frictional force
Asnwer (a) The force is in the same direction of the motion. Work done by the pulling force,
W = F × s = (10)(2) = 20J
(b) The force is not in the same direction of motion, work done by the frictional force
W = F × s × cos180o= (5)(2)(-1) = -10J
Work Done Against the Force of Gravity
Physics Department SMK Sultan Ismail Johor Bahru
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Example 3
Ranjit runs up a staircase of 35 steps. Each steps is 15cm in height. Given that Ranjit's mass
is 45kg, find the work done by Ranjit to reach the top of the staircase.
Answer In this case, Ranjit does work to overcome the gravity.
Ranjit's mass = 45kg
Vertical height of the motion, h = 35 × 0.15
Gravitational field strength, g = 10 ms-2
Work done, W = ?
W = mgh = (45)(10)(35 × 0.15) = 2362.5J
Force - Displacement Graph
In a Force-Displacement graph, work done is equal to the area in between the graph and the
horizontal axis.
Example 4
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The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m.
Find the work done by the force to move the trolley.
Answer
In a Force-Displacement graph, work done is equal to the area below the graph. Therefore,
work done
Energy Energy is defined as the capacity to do work.
Work is done when energy is converted from one form to another.
Unit: Nm or Joule(J)
Kinetic Energy Kinetic energy is the energy of motion.
Example 5
Determine the kinetic energy of a 2000-kg bus that is moving with a speed of 35.0 m/s.
Answer:
Kinetic Energy, Ek = ½ mv2
Physics Department SMK Sultan Ismail Johor Bahru
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Gravitational Potential Energy Gravitational potential energy is the energy stored in an object as the result of its vertical
position (i.e., height).
Formula:
Example 6
A ball of 1kg mass is droppped from a height of 4m. What is the maximum kinetic energy
possessed by the ball before it reached the ground?
Answer According to the principle of conservation of energy, the amount of potential energy losses is
equal to the amount of kinetic energy gain.
Maximum kinetic energy
= Maximum potentila energy losses
= mgh = (1)(10)(4) = 40J
Elastic Potential Energy Elastic potential energy is the energy stored in elastic materials as the result of their
stretching or compressing.
Formula:
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Example 7
Diagram above shows a spring with a load of mass 0.5kg. The extention of the spring is 6cm,
find the energy stored in the spring.
Answer
The energy stored in the spring is the elastics potential energy.
Conservation of Energy and Work Done During a conversion of energy,
Amount of Work Done = Amount of Energy Converted
Example 8
A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and it stops
at B after 4 seconds. What is the work done to overcome the friction?
Answer
In this case, kinetic energy is converted into heat energy due to the friction. The work done to
overcome the friction is equal to the amount of kinetic energy converted into heat energy,
hence
Power Power is the rate at which work is done, which means how fast a work is done.
Formula:
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Example 1
An electric motor takes 20 s to lift a box of mass 20kg to a height of 1.5 m. Find the amount
of work done by the machine and hence find the power of the electric motor.
Answer Work done,
W = mgh = (20)(10)(1.5) = 300J
Power, P =
Efficiency The efficiency of a device is defined as the percentage of the energy input that is transformed
into useful energy.
In the example above, the input power is 100J/s, the desire output power (useful energy) is
only 75J/s, the remaining power is lost as undisire output. Therefore, the efficiency of this
machine is
75/100 x 100% = 75%
Physics Department SMK Sultan Ismail Johor Bahru
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Elasticity Elasticity is the ability of a sub-stance to recover its original shape and size after distortion.
Forces Between Atoms
The intermolecular forces consist of an attractive force and a repulsive force.
At the equilibrium distance d, the attractive force equal to the repulsive force.
If the 2 atoms are brought closer, the repulsive force will dominate, produces a net
repulsive force between the atoms.
If the 2 atoms are brought furhter, the attractive force will dominate, produces a net
attractive force between the atoms.
Graph of Forces Between 2 atoms
X0 = Equilibrium Distance
When the particles are compressed, x < x0, the repulsive force between the particles
increases.
When the particles are stressed, x > x0, the attractive force between the particles increases.
If the distance x exceeds the elastic limit, the attractive force will decreases.
Physics Department SMK Sultan Ismail Johor Bahru
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Hooke's Law Hooke's Law states that if a spring is not stretched beyond its elastic limit, the force that
acts on it is directly proportional to the extension of the spring.
Elastic Limit The elastic limit of a spring is defined as the maximum force that can be applied to a spring
such that the spring will be able to be restored to its original length when the force is
removed.
Equation derived from Hooke's Law From Hook's Law, we can derived that
Spring Constant Spring constant is defined as the ratio of the force applied on a spring to the extension of the
spring.
It is a measure of the stiffness of a spring or elastic object.
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Graph of Streching Force - Extension
Gradient = Spring constant
Area below the graph = Work done
F-x graph and spring constant
The higher the gradient, the greater the spring constant and the harder (stiffer) spring.
For example, the stiffness of spring A is greater than spring B.
System of Spring
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Arrangement in series:
Extension = x × number of spring
Stiffness decreases
Spring constant = k/number of spring
Arrangement in parallel:
Extension = x ÷ number of spring
Stiffness increases
Spring constant = k × number of spring
Factors Affecting the Stiffness of Spring
Stiffer Less stiff
Material type of spring
Diameter of wire of
spring
Diameter of the spring
Length of the string