Work: In physics, a force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force. The work W done by a constant force of magnitude F on a point that moves a displacement s in a straight line in the direction of the force is the product W = Fs.
Work: In physics, a force is said to do work
if, when acting, there is a displacement of the point of application in the direction of the
force. The work W done by a constant
force of magnitude F on a point that
moves a displacement s in a straight line
in the direction of the force is the product W = Fs.
For example, if a force of 10 newtons (F = 10 N) acts along a point that travels 2
meters (s = 2 m), then
it does the work W = (10 N) (2 m) = 20 N m = 20 J. This is
approximately the work done lifting a 1 kg weight from ground level to
over a person's head against the force of gravity. Notice that the work is
doubled either by lifting twice the weight the same distance or by lifting the
same weight twice the distance.
The
small amount of work δW that occurs over an instant of time dt is
calculated as
(a) (b)
(c)
Fig. 44
This calculation can be generalized for a
constant force that is not directed along the line, followed by the particle
(Fig. 44b,c).
where θ
is the angle between the force vector and the direction of movement
Positive and Negative Work:
Work
done can be zero, positive or negative (Fig. 45a). When 0o <= θ
< 90o, work done is positive as Cosθ is positive. Work done by a force is positive if the applied
force has a component in the direction of the displacement. When a body is
falling down, the force of gravitation is acting in the downward direction. The
displacement is also in the downward direction. Thus the work done by the
gravitational force on the body is positive (Fig. 45b). Consider the same body
being lifted in the upward direction. In this case, the force of gravity is
acting in the downward direction. But, the displacement of the body is in the
upward direction. Since the angle between the force and displacement is 180o,
the work done by the gravitational force on the body is negative (Fig. 45c).
Note, that in this case the work done by the
applied force which is lifting the body up is positive since the angle between
the applied force and displacement is positive.
Similarly, frictional force is always
opposing the relative motion of the body. When a body is dragged along a rough
surface, the frictional force will be acting in the direction opposite to the
displacement. The angle between the frictional force and the displacement of
the body will be 180o. Thus, the work done by the frictional force
will be negative (Fig. 45d).
(a) (b)
(c)
(d)
Fig. 45
Energy: In physics, energy is the capacity for doing work. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work, i.e., energy in the process of transfer from one body to another. After it has been transferred, energy is always designated according to its nature. Hence, heat transferred may become thermal energy, while work done may manifest itself in the form of mechanical energy. Then mechanical energy can be converted to electrical energy (Fig. 46a). Fig. 46b shows the conversion of kinetic energy and potential energy.
(a) (b)
Fig. 46
Power: It is the rate of doing work. In other words, the amount of energy consumed per unit time. It is a scalar quantity. In the SI system, the unit of power is the joule per second (J/s), known as the watt. Fig. 47 defines the horsepower (hp). The equation for power can be written:
Fig. 47
Problem 10: A car of mass 500 kg is travelling along a horizontal
road. The engine of the car is working at a constant rate of 5 kW. The total
resistance to motion is constant and is 250 N. What is the acceleration of the
car when its speed is 5 m/s?
Solution: The equation of motion horizontally (from Newton's Second Law):
|
Fig. 48 |
Work-Energy
Principle:
The
principle of work and kinetic energy (also known as the work–energy principle) states that the work done by all forces
acting on a particle (the work of the resultant force) equals the change in the
kinetic energy of the particle. That is, the work W done by the resultant force on a particle equals the change in the particle's kinetic
energy Ek.
where
v1 and v2 are the speeds of the particle before and
after the work is done and m is its mass.
In the case, the resultant force F is
constant in both magnitude and direction, and parallel to the velocity of the
particle, the particle is moving with constant acceleration a along a
straight line. The relation between the net force and the acceleration is given
by the equation F = ma (Newton's second law), and the particle displacement s can be expressed by the equation
The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains:
Fig. 49
Problem 11: A 1 kg brick is
dropped from a height of 10 m. Calculate the work that has been done on the
brick between the moment it is released and the moment when it hits the ground.
Assume that air resistance can be neglected.
Solution: Potential energy of the brick at the height of 10 m, Ep = m. g. h = 1 × 9.8 × 10 = 98 J.
The brick had 98 J of potential energy when
it was released and 0 J of kinetic energy. When the brick hit the ground, it
had 0 J of potential energy and 98 J of kinetic energy. Therefore, Ek,i = 0 J and Ek,f = 98 J.
From the work-energy theorem: Wnet
= ΔEk = Ek,f − Ek,i = 98
− 0 = 98 J
Hence, 98 J of work was done on the brick.
The Impulse-Momentum
Change Theorem:
The
impulse-momentum theorem states that the change in momentum of an object equals
the impulse applied to it.
J = Δp
Newton's second law:
The equation indicates: Impulse = Change in momentum
∴ Impulse = (Final momentum) – (Initial momentum)
Impulse
Affects Momentum:
Any
moving object can have momentum.
This is because momentum is mass in motion. The way we determine an object's
momentum is fairly straightforward. Momentum is the object's mass times its velocity, or, in equation form, p=mv, where p is momentum, m is mass in
kilograms, and v is velocity in meters per second. Momentum is
proportional to both mass and velocity, meaning that a change in one will cause
the same amount of change in the other. So if you increase an object's mass,
you also increase its momentum. The same is true for velocity: increase or
decrease the object's speed, and you increase or decrease its momentum by the
same amount.
But
usually the object's velocity changes instead of its mass. A change in velocity
means the object is accelerating. Acceleration
is caused by a force and that the greater the force, the greater the
acceleration. Therefore, the greater the acceleration, the greater the
momentum.
Force
is an important factor, but time also counts. Specifically, when we are
interested in knowing how long the force acts. For example, if you push a box
across the floor for just a few seconds, the time interval is very short. But
if you push a box across the floor and you do so with the same force as before,
but this time for several minutes, you've increased the amount of time the
force acts. This longer time interval leads to a greater change in momentum.
This change in momentum is called impulse,
and it describes the quantity that we just saw: the force times the time
interval it acts over. The greater the impulse, the greater the change in
momentum. To change the impulse, you can either change the amount of force, or
you can change the time interval in which that force acts.
Problem 12: A ball with a
mass of 0.350 kg bounces off of a wall. Initially, it traveled horizontally to
the right, toward the wall at 25.0 m/s. In then bounces, and travels
horizontally to the left, away from the wall at 15.0 m/s. What is the impulse
of this collision between the ball and the wall?
Solution: The first step is to define a
positive direction. The problem can be solved with either the left or right
horizontal direction defined as positive, but for this solution, the positive
direction will be horizontally to the left (away from the wall). With this
definition, the initial velocity of the ball is v ⃗_2=+15.0m/s (horizontally), and the final velocity of the
ball is
The final momentum is:
The impulse of the collision is:
The impulse
of the collision is 14.0 N. s, horizontally to the left.
Exercise:
Chapter 1: Mechanics
Q1. When a body is said to
experience rectilinear motion?
Q2. Find the equations to
calculate the displacement, velocity, and acceleration.
Q3. For uniform circular
motion what would be the speed, velocity, acceleration and net force of a body?
Without this force what would be the direction of the moving body? When does a
moving body change its direction and undergoing an inward acceleration?
Q4. What are the fundamental
equations for the linear and circular motion?
Q5. Find the speed and
velocity from the figures.
Q6. Explain the direction of
angular acceleration from the figures.
Q7. Give examples about the
Newton’s laws of motion.
Q8. Relate the Polar and
Cartesian coordinates.
Q9. Compare non-inertial
reference frame with inertial reference frame.
Q10. Find the moment
(torque) from the figures.
Q11. How can you explain the
moment of inertia as an angular mass?
Q12. Derive the mass moment
of inertia as .
Q13. Derive the moment of inertia of a uniform rigid rod of length L and mass
M about an axis perpendicular to the rod and passing through O, at an arbitrary
distance h from one end.
Q14. Explain the radius of
gyration from the figure.
Q15. Write the use of polar
moment of inertia.
Q16. Show that for circular
section the polar moment of area is .
Q17. Give example for
positive and negative work.
Q18. Find the horsepower
(hp) from the figure.
Q19. Derive the expression
of Work-Energy Principle.
Q20. Explain the impulse-momentum theorem
from the figure.