physics 1501: lecture 12, pg 1 physics 1501: lecture 12 l announcements çhw 04 due this friday....

Physics 1501: Lecture 12, Pg 1

Physics 1501: Lecture 12Physics 1501: Lecture 12 Announcements

HW 04 due this Friday.Midterm 1: Monday Oct. 3Practice test on webHW solutions all this Friday.Office hours Friday instead of Wednesday

TopicsWork & EnergyPower


ReviewReviewDefinition of Work:Definition of Work:

Work (W) of a constant force FF acting through a displacement rr is:

W = FF . rr = F rr cos = Fr rr

Definition of Kinetic Energy :Definition of Kinetic Energy :The kinetic energy of an object of mass (m) moving at speed (v) is:

K = 1/2 m v1/2 m v22

Work Kinetic-Energy Theorem:Work Kinetic-Energy Theorem:


Lecture 12, Lecture 12, ACT 1ACT 1Work & EnergyWork & Energy

Two blocks having mass m1 and m2 where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. > 0) which slows them down to a stop.Which one will go farther before stopping ?

(a)(a) m1 (b)(b) m2 (c)(c) they will go the same distance

m1

m2



Hint: How much work does friction do on each block ??

(a)(a) m1 (b)(b) m2 (c)(c) they will go the same distance

m1

m2


Lecture 12, Lecture 12, ACT 1ACT 1SolutionSolution

The work-energy theorem says that for any object WWNETNET = = KK In this example the only force that does work is ffriction (since both N and mg are perpendicular to the blocks

motion).

mf

N

mg


Lecture 12, Lecture 12, ACT 1ACT 1 SolutionSolution

The work-energy theorem says that for any object WWNETNET = = KK In this example the only force that does work is ffriction (since

both N and mg are perpendicular to the blocks motion). The net work done to stop the box is - f D = -mg D.

m

D

This work “removes” the kinetic energy that the box had: WNET = KFIN - KINIT = 0 - KINIT


Lecture 12, Lecture 12, ACT 1ACT 1 SolutionSolution

The net work done to stop a box is - f D = -mg D.This work “removes” the kinetic energy that the box had:WNET = KFIN - KINIT = 0 - KINIT

This is the same for both boxes (same starting kinetic energy).

m2g D2m1g D1 m2 D2m1 D1

m1

D1

m2

D2

Since m1 > m2 we can see that D2 > D1



You like to drive home fast, slam on your brakes at the bottom of the driveway, and screech to a stop laying rubber all the way. It’s particularly fun when your mother is in the car with you. You practice this trick driving at 20 mph and with some groceries in your car with the same mass as your mama. You find that you only travel half way up the driveway. Thus when your mom joins you in the car, you try it driving twice as fast. How far will you go this time ?(a)(a) The same distance. Not so exciting.The same distance. Not so exciting.

(b) 2 times as far (only 7/10 of the way up the driveway)

(c)(c) twice as far, right to the door. Whoopee!twice as far, right to the door. Whoopee!

(d)(d) four times as far. Crashes into house. Sorry Ma.four times as far. Crashes into house. Sorry Ma.



Kinetic energy initially is K = 1/2 mv2

Work done (to get to a stop) is W = -fd

NOTE: force of friction, f=N is same in both cases, only v changes

Answer (e) 4


Work & Power:Work & Power:

Two cars go up a hill, a BMW Z3 and me in my old Mazda GLC. Both have the same mass.

Assuming identical friction, both engines do the same amount of work to get up the hill.

Are the cars essentially the same ? NO. The Z3 gets up the hill quicker It has a more powerful engine.



Power is the rate at which work is done. Average Power is,

Instantaneous Power is,



Consider the following,

But,

So,

Z3

GLC


Lecture 12,Lecture 12, ACT 3ACT 3Work & PowerWork & Power

Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. The instantaneous power delivered by the engine during this drive looks like which of the following,

A)

B)

C)Z3

timePow

erP

owe r

Pow

e r

time

time


Lecture 12,Lecture 12, ACT 3ACT 3SolutionSolution

We know that P = F.v Since there is constant acceleration, there must be a constant

force (ma) from the engine to counteract gravity, mgcos. Constant acceleration also means that v = v0 + at So P = ma.at = ma2 t At the top P = 0

Answer is A)

Z3

timePow

er


Work & Power:Work & Power: Power is the rate at which work is done.

InstantaneousPower:

AveragePower:

A person of mass 80.0 kg walks up to 3rd floor (12.0m). If he/she climbs in 20.0 sec what is the average power used.

W = F h = (mg) h

W = 80.0kg 9.8m/s2 12.0 m = 9408 J

W = Fx dxP = W / t

P = W / t = 9408 J / 20.0s = 470 W

Simple Example 1 :Simple Example 1 :

Units (SI) areWatts (W):

1 W = 1 J / 1s



Engine of a jet develops a trust of 15,000 N when plane is flying at 300 m/s. What is the horsepower of the engine ?

Simple Example 2 :Simple Example 2 :

P = F vP = (15,000 N) (300 m/s) = 4.5 x 106 W = (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp !


Lecture 12,Lecture 12, ACT 4ACT 4Power for Circular MotionPower for Circular Motion

I swing a sling shot over my head. The tension in the rope keeps the shot moving in a circle. How much power must be provided by me, through the rope tension, to keep the shot in circular motion ?

Note that Rope Length = 1m

Shot Mass = 1 kg

Angular frequency = 2 rads/sec v

A) 16 J/s B) 8 J/s C) 4 J/s D) 0



Before calculating anything think the problem through and draw a diagram.

v

D) 0

• We know that Power is F.v

• Note that F v

• Thus, P = Fv cos = 0.

T



Note that the string expends no power, i.e. does no work. Makes sense ?

By the work – kinetic energy theorem, work done equals change in kinetic energy.

K = 1/2 mv2, thus since v doesn’t change, neither does K.

A force perpendicular to the direction of motion does not change speed, v, and does no work.

v

T


Work Done Against GravityWork Done Against Gravity

Consider lifting a box onto the tail gate of a truck.

m

h

The work required for this task is,W = F · d = (mg)(h) (1) W = mgh


Work Done Against GravityWork Done Against Gravity Now use a ramp to help you with the task.

Is less work needed to get the box into the truck? (It’s “easier” to lift the box)

m

h


Work Done Against GravityWork Done Against Gravity

m

mg

mgsin

mgcos

F

To push the box with constant speed, F = mgsinThe length of the ramp is h/sinSo the work done is,

W = Fd = (mgsin)(h/sin)W = mgh

Same as before !

N

h


Work Done by a SpringWork Done by a Spring

Force from the spring is Fs = -kx,Displacement is x.

W = F dx = - kx dxW = - 1/2 kx2

Remember these two results for Chapter 7.

x

Fs

physics 1501: lecture 12, pg 1 physics 1501: lecture 12 l announcements çhw 04 due this friday....

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