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Sect. 7-4: Kinetic Energy; Work-Energy Principle

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- Energy: Traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition.
- Kinetic Energy The energy of motion
“Kinetic” Greek word for motion

An object in motion has the ability to do work.

- Consider an object moving in straight line. Starts at speed v1. Due to the presence of a net force Fnet, it accelerates (uniformly) to speed v2, over distance d.
Newton’s 2nd Law: Fnet= ma (1)

1d motion, constant a (v2)2 = (v1)2 + 2ad

a = [(v2)2 - (v1)2]/(2d) (2)

Work done: Wnet = Fnet d (3)

Combine (1), (2), (3):

- Fnet= ma (1)
- a = [(v2)2 - (v1)2]/(2d) (2)
- Wnet = Fnet d (3)
Combine (1), (2), (3):

Wnet = mad = md [(v2)2 - (v1)2]/(2d)

or

Wnet = (½)m(v2)2 – (½)m(v1)2

- Summary:The net work done by a constant force in accelerating an object of mass m from v1to v2is:
DEFINITION:Kinetic Energy (KE)

(for translational motion; Kinetic = “motion”)

(units are Joules, J)

- We’ve shown:The WORK-ENERGY PRINCIPLE
Wnet = K( = “change in”)

We’ve shown this for a 1d constant force. However, it is valid in general!

The net work on an object = The change in K.

Wnet = K

The Work-Energy Principle

Note!:Wnet = work done by the net (total) force.

Wnetis a scalar & can be positive or negative (because K can be both + & -).If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases.

The SI Units are Joulesfor both work & kinetic energy.

The Work-Energy Principle

Wnet = K

NOTE!

This isNewton’s 2nd Lawin

Work & Energy Language!

Table from another textbook

A moving hammer can do work on a nail!

For the hammer:

Wh = Kh = -Fd

= 0 – (½)mh(vh)2

For the nail:

Wn = Kn = Fd

= (½)mn(vn)2 - 0

Example 7-7: Kinetic energy & work done on a baseball

A baseball, mass m = 145 g(0.145 kg) is thrown so that it acquires a speed v = 25 m/s.

a. What is its kinetic energy?

b. What was the net work done on the ball to make it reach this speed, if it started from rest?

Calculate the net work required to accelerate a car, mass m = 1000-kg, from v1 = 20 m/s to v2 = 30 m/s.

A car traveling at speedv1 = 60 km/hcan brake to a stop within a distanced = 20 m. If the car is going twice as fast,120 km/h, what is its stopping distance? Assume that the maximum braking force is approximately independent of speed.

Wnet = Fd cos (180º) = -Fd(from the definition of work)

Wnet = K = (½)m(v2)2 – (½)m(v1)2(Work-Energy Principle)

but, (v2)2 = 0 (the car has stopped) so-Fd = K= 0 - (½)m(v1)2

or d (v1)2

So the stopping distance is proportional to the square of the initial speed! If the initial speed is doubled, the stopping distance quadruples!

Note:K (½)mv2 0Must be positive, since m & v2 are always positive (real v).

Example 7-10: A compressed spring

A horizontal spring has spring constant k = 360 N/m. Ignore friction.

a. Calculate the work required to compress it from its relaxed length (x = 0) to x = 11.0cm.

b. A1.85-kg block is put against the spring. The spring is released. Calculate the block’s speed as it separates from the spring at x = 0.

c. Repeat part b. but assume that the block is moving on a table & that some kind of constant drag force FD = 7.0 N(such as friction) is acting to slow it down.

Example

A block of mass m = 6 kg, is pulled

from rest (v0 = 0) to the right by a

constant horizontal force F = 12 N.

After it has been pulled for Δx = 3 m,

find it’s final speed v.

Work-Kinetic Energy Theorem

Wnet = K (½)[m(v)2 - m(v0)2] (1)

If F = 12 N is the only horizontal force,

then Wnet = FΔx (2)

Combine (1) & (2):

FΔx = (½)[m(v)2 - 0]

Solve for v: (v)2 = [2Δx/m]

(v) = [2Δx/m]½ = 3.5 m/s

FN

v0

Conceptual Example

A man wants to load a refrigerator onto a truck bed

using a ramp of length L, as in the figure. He claims

that less work would be required if the length L were

increased. Is he correct?

A man wants to load a refrigerator

onto a truck bed using a ramp of

length L. He claims that less work

would be required if the length L

were increased. Is he correct?

NO!For simplicity, assume that it is wheeled up the on a dolly atconstant speed. So the kinetic energy change from the ground to the truck isK = 0. The total work done on the refrigerator is Wnet = Wman + Wgravity + Wnormal

The normal forceFNon the refrigerator from the ramp is at90º to the horizontal displacement & does no work on the refrigerator (Wnormal = 0).Since K = 0, by the work energy principle the total work done on the refrigerator is Wnet = 0. = Wman + Wgravity. So, the work done by the man isWman = - Wgravity.The work done by gravity isWgravity = - mgh[angle between mg & h is 180º & cos(180º) = -1]. So, Wman = mgh

No matter what he does he still must do the

SAME amount of work

(assuming height h = constant!)