Download 1 Saturday X Saturday X-tra X-Sheet 6 Work

Saturday XX-tra
X-Sheet 6
Work - Energy
Key Concepts
This lesson focuses on the following:
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Defining work
Components of force and work
Kinetic and Potential energy
Mechanical energy
Work-Energy Theorem
Terminology & definitions
• Work – the product of the force, and the displacement covered by an object in the direction of the
force.
• Energy – the ability to do work.
• Gravitational potential energy – energy possessed by an object due to its position above the earth.
• Kinetic energy – energy possessed by an object due to its motion.
• Mechanical energy – the sum of gravitational potential energy and kinetic energy of an object.
X-planation of key concepts and terminologies
Work is defined as the product of the force and the straight line distance covered by an object in the
direction of the force. Mathematically: W = F
F∆x
∆x. The displacement and the force acting on an object
have to be in the same direction for work to be done. If a force acts at an angle, θ, to the horizontal
and yet the object moves in the horizontal direction, the horizontal component of the force needs to
be used as the force that caused the movement. Thus W = F∆xcosθ. When a force acts at an angle to
the horizontal causing the object to move along the horizontal direction, there is also a vertical
component of the force that tends to lift the object upwards. Such vertical component equals F∆xsinθ.
Work is not done when …
- an object does not move although a force is applied.
- when the force is at right angle (perpendicular) to the displacement.
- when there is no resultant force acting on the object the whole time.
Energy is the ability to do work. Energy is found in different forms, e.g., heat, light, sound, etc. When
work is done on a system, energy is transferred from one form to another. Energy and work only
have magnitudes and no direction and, therefore, they are both scalar quantities.
Gravitational potential energy is the energy possessed by an object due to its position above the
surface of the earth. It is found by using the formula, Ep = mgh, where:
m is the mass of the object,
g is the acceleration due to gravity = 9,8 m·s-2,
h is the vertical height (position of the object above the surface of the earth.
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Kinetic energy is the energy possessed by an object due to its motion. It is found by using the
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formula, E k = mv 2 where; m is the mass of the object and v is the velocity of the object.
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The sum of kinetic energy and gravitational potential energy for an object is called mechanical
energy: that is ME = Ep + Ek. Mechanical energy is conserved in a closed system: that is Ep + Ek =
constant in a closed system.
The work–energy theorem states that the total work done on an object equals the change in the
kinetic energy of an object (Wnet = ∆Ek). In a situation where the motion is along an incline surface
with friction, there is work done against friction, and work is also done in lifting the object from the
bottom of the ramp to the top. In this case Wnet = ∆Ek + ∆Ep - Wf.
X-ample Questions
Questions
1.
1.1 An object experiences a force of 20 N along the horizontal and covers a
displacement of 10 m.
Find the work done by the force on the object.
1.2 An object of mass 2 kg accelerates at 3 m·s-2 and covers a displacement of 10 m.
Calculate the
work done on the object.
2. A 100 N force acts on an object at an angle of 600 to the horizontal. The object
horizontal distance of 5 m.
2.1 How much work is done by the applied force on the object?
2.2 Calculate the magnitude of the force that tends to lift the object.
then moves a
3. A block of mass 500 g is lifted to a vertical distance of 80 m. Calculate the amount of
gravitational potential energy gained by the block.
4.
An object of mass 450 g is moving at a speed of 10 m•s-1. How much kinetic
energy does the object have?
5. A builder standing on the roof of a building allows a hammer of mass 2 kg to fall
height of 10 m above the ground.
5.1 Calculate the gravitational potential energy of the hammer before it was released
builder.
5.2 Use the law of conservation of mechanical energy to calculate the kinetic energy of
a height of 5 m above the ground during its fall.
5.3 What is the total kinetic energy of the hammer just before it strikes the ground?
answer.
freely from a
by the
he hammer at
Explain your
2
6.
A person skis down a 20 m long snow slope which makes an angle of 25° with the horizontal.
The total mass of the skier and skis is 50 kg. There is a constant frictional force of 60 N
opposing the skier's motion. The speed of the skier as he/she descends from the top of the
slope is 2,5 m·s-1
vi = 2,5
m·s-1
20 m
25
o
6.1
Calculate the magnitude of the net force parallel to the slope experienced by the person.
6.2
Calculate the maximum speed of the skier at the bottom of the 20 m slope
(5)
(6)
X-ercise
1.
A box of mass 60 kg starts from rest at height h and slides down a rough slope of length 10
m, which makes an angle of 25° with the horizontal. It undergoes a
constant acceleration of
-2
magnitude 2 m·s while sliding down the slope
60 kg
A
8m
h =1,2 m
25º
1.1
State the work-energy theorem in words.
1.2
Draw a free-body diagram to show ALL the forces acting on the cardboard box
while
it slides down the slope.
1.3
The box reaches the bottom of the slope.
Calculate the following:
1.3.1
The kinetic energy of the box, using the equations of motion.
1.3.2 The work done on the box by the gravitational force.
1.3.3 The work done on the box by the frictional force, using the work-energy
theorem.
1.3.4 The magnitude of the frictional force acting on the box.
Answers
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1.1
The net work done on an object is equal to the change in the
object's kinetic energy. OR/OF
The work done on an object by a net force is equal to the change in the object's
kinetic energy.
(2)
1.2
N/FN/Force of surface on object ●
f/Ff/Force of friction Fg/Force of earth on object OR/OF
N/FN/Force of surface on object ●
f/Ff/Force of friction F g⊥ Component of gravitational force perpendicular to incline
Fg// Component of gravitational force parallel to incline
(3)
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1.3.1
+
v 2f = v i2 + 2a∆x = (0)2 + (2)(2)(10) = 40 m2·s-2
Ekf = ½m v 2f = ½(60)(40) = 1 200 J OR/OF
∆x = v i ∆t + 21 a∆t 2
10 = (0)∆t ½(2) ∆t2
∴ ∆t = 3,16 s
vf = vi + a∆t = 0 + (2)(3,16) = 6,32 m·s-1
Ekf = ½m v 2f = ½(60)(6,32)2 = 1 200 J (5)
1.3.2
Wg = w// ∆xcosθ = mgsin 25° (10)(cos 0°)
= (60)(9,8)sin25°10(1)
= 2 485 J OR
Wg = w∆xcosθ = mghcos 0°
= (60)(9,8) (10)sin25°(1) = 2 485 J OR
Wg = -∆U = - (0 – mgh) = - (0 – (60)(9,8)(10)sin25° = 2 485 J 1.3.3
Wnet = ∆Ek
Wg(parallel to slope) + Wf = ∆Ek 2 485 + Wf = 1 200 Wf = - 1 285 J 1.3.4
Wf = Ff ∆xcosθ - 1 285 = f(10)cos180° Ff = 128,5 N (4)
(3)
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