Course description

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  (horizontally). The initial momentum is:

The final momentum is:

      (p_2 ) ⃗=+5.25 kg.m/s

The impulse of the collision is:

The impulse of the collision is 14.0 N. s, horizontally to the left.                                                                  Fig 50                                                                              

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.

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