Physics Work and Kinetic Energy Teacher: Luiz Izola

Physics

Work and Kinetic Energy

Teacher: Luiz Izola

Chapter Preview

1. Work Done by a Constant Force

2. Kinetic Energy

3. Work-Energy Theorem

4. Work Done by A variable Force

5. Power

Introduction

We know by intuition that motion, energy, and work are somehow related.

Work can be thought as force times displacement.

In this chapter will show the relation between work and energy of motion (kinetic energy).

Work Done by a Constant Force

First, we will consider the case where force and displacement have the same direction.

Later, we will consider cases where force and displacement have arbitrary directions.

Then, we will learn how to calculate work done on an object which has several forces acting on it.

Force in the Direction of Displacement

The greater the force, the greater the work. The greater the distance, the greater the work.

Work = Force x Displacement Measured in Joules (J) (Newton x Meter) Work is a scalar. (Not a vector). 1J = kg. m2/s2

Force in the Direction of Displacement

Ex: A finch can exert a force of 205N with its beak as it cracks a seed case. If the beak moves 0.40cm, how much work does it do to get the seed?

Ex: An intern pushes a 72-kg patient on a 15-kg gurney, producing an acceleration of 0.60m/s2. How much work does the intern do by pushing the patient 2.50meters. Assume no friction.

Force at an Angle to Displacement

A person pulling a suitcase on a level surface makes an angle Ө with the horizontal.

Work is the component of force in the direction of the displacement times the displacement.

W = (FcosӨ)d = Fdcos(Ө). The unit is J.

What is the effect on work for Ө = 0o or 90o?


Work is the component of force in the direction of displacement times the magnitude of the displacement or the component of displacement in the direction of force times the magnitude of the force.

Remember that the result is a scalar. It is just the multiplication of the magnitudes of displacement and force vectors.


Ex: A 75-kg person slides 5.0-m on a water slide, dropping through a vertical line of 2.50-m. How much work does the gravity do on the person?


Ex: You want to load a box into a truck. One way is to lift the box straight up through a height “h”, doing a work W1. Otherwise, you can slide the box up a ramp a distance “L”, doing work W2. Which is correct: (a) W1 < W2 (b) W1 = W2 (c) W1 > W2

Negative Work and Total Work Work depends on the angle between force and displacement (or motion direction). • Work is positive if the force has a component in the direction of motion (Ө < 90o).• Work is zero if the force has no component in the direction of motion (Ө = 90o).• Work is negative if the force has a component in the direction of motion (Ө > 90o).

When more than one force acts on an object, the total work is the sum of the work done by each force separately or the work of the Ftotal.

Negative Work and Total Work

Ex: A car of mass “m” coasts down a hill inclined at an angle Ф below the horizontal. The car is acted on by 3 forces: (a) The normal force N applied by the road, (b) The force due to air resistance Fair, and (c) The gravity force mg. Find the total work done on the car as it travels a distance “d” along the road.

Negative Work and Total Work

Ex: Calculate the total work from the previous example by using Wtotal = FtotaldcosФ.

Kinetic Energy and Work-Energy Theorem

If the total work done in an object is positive, its speed increases.

If the total work done in an object is negative, its speed decreases.

Knowing that Kinetic Energy, K, is defined as:

K = 1/2mv2, where: m = mass

v = velocity.

The unit is (kg)(m2/s2) = J (Joule)


Ex: A truck moving at 15m/s has a kinetic energy of 1.4x105 J. What is the mass of the truck?

Work-Energy Theorem

“The total work done on an object is equal to the change in its kinetic energy.”

Wtotal = ΔK = 1/2mvf2 – 1/2mvi2

Ex: How much work is required for a 74-kg sprinter to accelerate from rest to 2.2m/s?


Ex: A 4.1-kg box of books is lifted vertically from rest a distance of 1.60m by an upward force of 60.0N. Find (a) Work done by the applied force. (b) The work done by gravity. (c) The final speed of the box.


Ex: A boy exerts a force of 11.0N at 29.0o above the horizontal on a 6.40-kg sled. Find the work done by the boy and the final speed of the sled as it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction.

Work Done by a Variable Force

If a force varies continuously with position, we can approximate it with a series of constant values that follow the shape of a curve.

The work done by the continuous force is approximately equal to the sum of the rectangles’ areas.

In the limit of an infinite number of very small rectangles, the total area of all rectangles becomes identical to the area under the curve.



A particular case is the work to stretch/compress a spring a distance x from equilibrium

W = 1/2kx2 (k measures in N/m)


Ex: (a) If the work required to stretch the Slinky Dog 1 meter is 2J, what is the force constant for the Slinky Dog? (b) How much work is required to stretch it from 1m to 2m?

Power Power is a measure of how quick the work is

done. Therefore, power is defined as:

P = Work (W) / Time (t) = W/t = Fv

The International Unit is: (watts) W = J/s

Ex: To pass a truck, your 1.30x103kg car needs to accelerate from 13.4m/s to 17.9m/s in 3.00s. What is the minimum power required to pass it?

Power

Ex: It takes a force of 1280N to keep a 1500-kg car moving with constant speed up to a slope of 5.0o. If the engine delivers 50 hp (50x746W) to the drive wheels, what is the maximum speed of the car?

Ex: Calculate the power output of a 1.3g fly as it walks straight up a window pane at 2.5cm/s.

HomeworkPage 193-194: #1, #3, #5, #7, #9, #11, #13

Page 194-195: #15, #17, #18, #21, #22, #25, #27, #29, #31

Page 195: #33, #35, #37, #39, #41, #43, #45, #47

Documents

Physics Work and Kinetic Energy Teacher: Luiz Izola