Home › 2020 › April › Page 62

Monthly Archives: April 2020

2.11.4 Gravitational Potential Energy

Posted on by

Gravitational potential energy is the energy stored in an object as the result of its vertical position (i.e., height).

Formula:


Example:
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

 

2.11.2 Energy

Posted on by
  1. Energy is defined as the capacity to do work.
  2. Work is done when energy is converted from one form to another.
  3. The unit of is the same as the unit of work, which is Nm or Joule (J)
  4. In SPM, we need to know the calculation related to the kinetic energy and potential energy.

 

2.11.1 Work

Posted on by
  1. Work done by a constant force is given by the product of the force and the distance moved in the direction of the force.
  2. The unit of Nm(Newton metre) or J(Joule).
  3. Work is a scalar quantity.

Formula




When the direction of force and motion are same, θ = 0o, therefore cosθ = 1
Work done,

W = F × s

Example:

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


Example:


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

  1. the work done by the pulling force
  2. the work done by the frictional force


Answer:
(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 × cos180°
W = (5)(2)(-1)
W = -10J


Work Done Against the Force of Gravity


Example:
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

Finding Work from Force-Displacement Graph

In a Force-Displacement graph, work done is equal to the area in between the graph and the horizontal axis.

Example:

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


 

2.10.3 Vectors in Equilibrium

Posted on by

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.
When the forces acting on an object are balanced, they cancel each other out. The net force is zero.

Effect :
  • an object at rest is continuely at rest [ velocity = 0]
  • a moving object will move at constant velocity [ a = 0]

Example:

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:

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

 

2.10.2 Vector Resolution

Posted on by


A vector can be resolve into 2 component which is perpendicular to each others.

Example:

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 cos30° =2598N
Friction = 500N

Resultant horizontal force = 2598N - 500N =2098N

Example:

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 = 25cos45° + 25cos45° = 35.36N

Vertical component of the forces = 25sin45° - 25sin45° = 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


Weight component along the plane = Wsinθ.
Weight component perpendicular to the plane = Wcosθ.

Example:

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 = 20sin30° = 10N

The resultant force along the plane = 20 - 2 - 10 = 8N

 

2.10.1 Vector and Scalar Quantity

Posted on by
  1. A scalar quantity is a quantity which can be fully described by magnitude only.
  2. A vector quantity is a quantity which is fully described by both magnitude and direction.

Vector Diagram

  1. The arrow shows the direction of the vector.
  2. 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


If 2 vectors (a and b) are perpendicular to each others, the magnitude and direction of the resultant vector can be determined by the following equation.


Example:
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:
Magnitude of the resultant force



Example:

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,

 

2.9.3 Apparent Weight in a Lift

Posted on by

  1. When a man standing inside an elevator, there are two forces acting on him.
    1. His weight, (W) which acting downward.
    2. Normal reaction (R), acting in the opposite direction of weight.
  2. The reading of the balance is equal to the normal reaction (R).
  3. Figure below shows the formula to calculate the reading of the balance at different situation.



Example 1:
Subra is standing on a balance inside an elevator. If Subra’s mass is 63kg, find the reading of the balance when the lift,
  1. stationary
  2. moving upward with a constant velocity, 15 ms-1.,
  3. moving upward with a constant acceleration, 1 ms-2.
  4. moving downward with a constant acceleration, 2 ms-2.

Answer:
a.

W = mg
W = (63)(10) = 630N
b.

W = mg
W = (63)(10) = 630N

c.

R = mg + ma
R = (63)(10) + (63)(1)
R = 693N
d.

R = mg - ma
R = (63)(10) - (63)(2)
R = 630 - 126 = 504N


Example 2:
A 54kg boy is standing in an elevator. Find the force on the boy's feet when the elevator
  1. stands still
  2. moves downward at a constant velocity of 3 m/s
  3. decelerates downward with at 4.0 m/s2, 
  4. decelerates upward withat 2.0 m/s2.

Answer:
a.

W = mg
W = (54)(10) = 540N
b.

W = mg
W = (54)(10) = 540N
c.

R = mg + ma
R = (54)(10) + (54)(4) = 756N

d.

R = mg - ma
R = (54)(10) - (54)(2) = 432N

 

2.9.2 Weight and Mass

Posted on by

Mass

  1. Mass is defined as the quantity of matter. It is a base quantity and also a scalar quantity, which mean it has no direction.
  2. The SI unit of mass is kilogram (kg)

Weight

  1. The weight of an object is defined as the gravitational force acting on the object.
  2. The SI unit of weight is Newton (N)

Differences between Weight and Mass

Weight Mass
Depends on the gravitational field strengthIndependent from the gravitational field strength
Vector quantityScalar Quantity
Unit Newton (N)Unit: Kilogram (kg)

 

 

2.9.1 Gravitational Field

Posted on by

A gravitational field as a region in which an object experiences a force due to gravitational attraction.

Gravitational Field Strength

  1. 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.
  2. The unit of gravitational field strength is N/kg.
  3. The gravitational field strength is denoted by the symbol "g".
    4. g = Gravitational Field Strength
    F = Force acted on an object
    m = mass of the object.

Gravitational Acceleration

  1. The gravitational acceleration is the acceleration of an object due to the pull of the gravitational force. It has the unit of ms-2
  2. The symbol of gravitational acceleration is " g ".
  3. Gravitational acceleration does not depend on the mass of the moving object.
  4. The magnitude of gravitational acceleration is taken to be 10ms-2.

Important notes:

  1. Gravitational acceleration does not depend on the mass of the moving object.
  2. The magnitude of gravitational acceleration is taken to be 10ms-2.

Gravitational Field Strength vs. Gravitational Acceleration

  1. Both the gravitational field strength and gravitational acceleration have the symbol, g and the same value (10ms-2) on the surface of the earth.
  2. When considering a body falling freely, the g is the gravitational acceleration.
  3. When considering objects at rest, g is the Earth’s gravitational field strength acting on it.

Free Falling

  1. Free falling is a motion under force of gravity as the only force acting on the moving object.
  2. Practically, free falling can only take place in vacuum.

Case of Free Falling 1 - Falling from High Place

When an object is released from a high place,

  1. its initial velocity, u = 0.
  2. its acceleration is equal to the gravitational acceleration, g, which taken to be 10ms-2 in SPM.
  3. the displacement is the of the object when it reaches the ground is equal to the initial height of the object, h.

Case of Free Falling 2 - Launching Object Upward

If an object is launched up vertically,

  1. the acceleration = -g (-10ms-2)
  2. the velocity become zero when the object reaches the highest point.
  3. the displacement of the object at highest point is equal to the vertical height of object, h
  4. the time taken for the object to move to the maximum height = the time taken for the object to fall from the maximum point to its initial position.