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PHYS 111N HW6: Work and energy
- Course
- PHYSICS
- Institution
- Old Dominion University
This is a comprehensive and detailed homework with correct solutions on Hw 6; Work and Energy.
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HW6: Work and energyDue: 11:59pm on Wednesday, April 4, 2018
You will receive no credit for items you complete after the assignment is due. Grading Policy
Understanding Work Done by a Constant Force
Learning Goal:
To explore the definition of work and learn how to find the work done by a force on an object.
The word "work" has many meanings when used in everyday life. However, in physics work has a very specific definition.
This definition is important to learn and understand. Work and energy are two of the most fundamental and important
concepts you will learn in your study of physics. Energy cannot be created or destroyed; it can only be transformed from one
form to another. How this energy is transferred affects our daily lives from microscopic processes, such as protein synthesis,
to macroscopic processes, such as the expansion of the universe!
When energy is transferred either to or away from an object by a force F acting over a displacement d , work
W is done on
that object. The amount of work done by a constant force can be found using the equation
W = F d cos θ ,
where F is the magnitude of F ,
d is the magnitude of d , and θ is the angle between F and
d.
The SI unit for work is the joule, J . A single joule of work is not very big. Your heart uses about 0.5 J each time it beats, and
the 60-watt lightbulb in your desk lamp uses 216, 000 J each hour. A joule is defined as follows:
2
kg m
1J = 1Nm = 1 s2
The net work done on an object is the sum of the work done by each individual force acting on that object. In other words,
Wnet = W1 + W 2 + W3 + ⋯ = ∑i W i .
The net work can also be expressed as the work done by the net force acting on an object, which can be represented by the
following equation:
Wnet = Fnet d cos θ.
Knowing the sign of the work done on an object is a crucial element to understanding work. Positive work indicates that an
object has been acted on by a force that tranfers energy to the object, thereby increasing the object's energy. Negative work
indicates that an object has been acted on by a force that has reduced the energy of the object.
The next few questions will ask you to determine the sign of the work done by the various forces acting on a box that is being
pushed across a rough floor. As illustrated in the figure , the box is being acted on by a normal force n , the force due to
force of kinetic friction
gravity w , the f k , and the pushing force F p . The displacement of the box is d .
,Part A
Which of the following statements accurately describes the sign of the work done on the box by the force of the push?
Hint 1. Find the angle
The work done on the box by the pushing force depends on the angle θ between F p and the displacement d .
What is this angle?
ANSWER:
0 degrees
45 degrees
90 degrees
180 degrees
ANSWER:
positive
negative
zero
Correct
Part B
, Which of the following statements accurately decribes the sign of the work done on the box by the normal force?
Hint 1. Finding theta
the displacement d . What
The work done on the box by the normal force depends on the angle θ between n and
is this angle?
ANSWER:
0 degrees
45 degrees
90 degrees
180 degrees
ANSWER:
positive
negative
zero
Correct
Part C
Which of the following statements accurately decribes the sign of the work done on the box by the force of kinetic
friction?
Hint 1. Finding theta
The work done on the box by the force of kinetic friction depends on the angle θ between f k and the
displacement d . What is this angle?
ANSWER:
0 degrees
45 degrees
90 degrees
180 degrees
, ANSWER:
positive
negative
zero
Correct
Part D
Which of the following statements accurately decribes the sign of the work done on the box by the force of gravity?
Hint 1. Finding the angle
The work done on the box by the weight depends on the angle θ between w and
the displacement d . What is
this angle?
ANSWER:
0 degrees
45 degrees
90 degrees
180 degrees
ANSWER:
positive
negative
zero
Correct
Making generalizations
You may have noticed that the force due to gravity and normal forces do no work on the box. Any force that is perpendicular
to the displacement of the object on which it acts does no work on the object.
The force of kinetic friction did negative work on the box. In other words, it took energy away from the box. Typically, this
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