Chapter 7 Work and Kinetic Energy

Which one costs energy?

Question: (try it) How to throw a baseball to give it large speed? Answer: Apply large force across a large distance!

Force exerted through a distance performs mechanical work. 1 Units of Chapter 7 • Work Done by a Constant Force • Kinetic Energy • The Work-Energy Theorem • Work Done by a Variable Force (optional) • Power

Read Chapter 8, Potential energy before the next lecture. We will finish Chapter 8 in the next lecture. 2 7-1 Work Done by a Constant Force When the force is parallel to the displacement: Constant force in direction of motion does work W.

(7-1)

SI unit: newton -meter (N·m) = Joule, J 1 J = 1 N.m

3 If F= 15 N, distance = 2 m , W=30 J 7-1 Work Done by a Constant Force

1J1 Jou le 1 J

How much is that?

4 If the force is at an angle to the motion, it does the following work:

(7-3) θ is the angle between force and motion direction.

Pulling at θ= 20 , F=15N, d=2 m 5 W= Fd cos θ =15*2* cos 20 = 28 J 7-1 Work Done by a Constant Force The work can al so b e writt en as th e d ot product of the force and the displacement:

θ is the angle between force and motion directi on.

6 The work done may be positive, zero, or negative, depending on the angle between the force and the motion:

Here for F and d we only use their sizes (absolute value). The sign of the work is determined only by the angle between that force and motion.

7 Special cases:

When force is When force is Opposite ppperpendicular to motion direction, to motion direction, cos(180)=-1. it does no work. Examples: Examples: Normal force is always Kinetic friction force perpendicular to surface . does negative work. W = – f d Tension of pendulum … fk k 8 Always! If there is more than one force acting on an object, we can find the work done by each force and add them together to find the total work.

(7-5)

Total work: Wtotal = W1 + W2 +W3+ ….. the sum of the work done by each forces. 9 Q: Is Work a scalar or a vector? Scalar! It on ly says how muc h energy i s add ed or used , (positive or negative), but doesn’t indicate motion directions or force directions. When we add work to get total work, we add as ppgpositive and negative simple numbers. We don’t add work as vector arrows.

x fk Fpull

If FPull= fk ; a=0 , v=constant WPull= FPull d ; Wfk= – fk d ; Wtotal = 0

If F Pull >f> fk ; a >0 , If v >0, v w ill increase. Wtotal = FPull d – fkd = (FPull –fk) d = Fnet d 10 7-2 Kinetic Energy and the Work-Energy Theorem When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.

11 7-2 Kinetic Energy and the Work-Energy Theorem

After algebraic manipulations of the equations of motion, we find: The total work done to one object is always equal to the change of its ½mv2

Therefore, we define the kinetic energy: (7-6)

Kinetic energy has SI unit: J (same dimension as Work) 1 kg m2/s2 = 1 (kg m/s2)m =1Nm =1 Joule 12 7-2 Kinetic Energy and the Work-Energy Theorem

Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy.

(7-7)

It’s true for ALL MOTIONS, no only for constant a motion!!! 13 m=1000kg, v0=0, µk=0.2 Problem solving strategy : (work problems) ο 1C1.Compu te wor kfk for in diiddividua lfl forces fi rst . φ=30 , d05d=0.5m, FidFind vf 2. Add all work together as scalar numbers. 3. Set equation mgg()= 9800 (N) N=mgcosφ If you know total work, you can solve v, fk=µkN=µkmgcosφ if you know v, you can solve Work. fk=0. 2∗9800∗cos30=1697(Ν)

Wmg=mg d cos 60 = 9800*0.5* cos60=2450 J Wfk = – fk d = – 1697*0.5= – 849 J ; WN=0 ; Wtotal = 2450 – 849 = 1601 J 14 K = K +W 2 f i total ½mvf =Wtotal =1601J vf=1.79m/s Wmg=mg d cos 60 =2450 J Wfk = – fk d = – 849 J ; W = 2450 + (– 849) = 1601 J Work is a scalar. total TtttlkdbllfTo get total work done by all forces, add workik in Joules directly as simple numbers! You only need to worry about the angles between each force and actual motion when you calculate work done by each force. After the work is calculated. Add them as simple numbers, no worry about direction any more. 15 Spring Force: Hooke’s Law Fspring= - k ∆x k :Hooke’s constant (how strong a spring is) ∆x : distance of stretch/compression Force direction: Always in the opposite direction of ∆ x Spring force always tries to recover its natural Length

Atten tion: This k is not K, Spring constant is not Kinetic 16 Energy. 7-3 Work Done by a Variable Force If the force is constant, we can interpret the work done graphically:

17 7-3 Work Done by a Variable Force

If the force takes on several successive constant values:

18 7-3 Work Done by a Variable Force

We can then approximate a continuously varying force by a succession of constant values.

19 7-3 Work Done by a Variable Force

The force needed to stretch a spring an amount x is F = kx. Therefore, the work done in stretching the spring is

(7-8)

20 7-4 Power

Power is a measure of the rate at which work is done:

(7-10)

SI unit : J/ s = watt , W 1 horsepower = 1 hp = 746 W

21 7-4 Power

22 7-4 Power

If an object is moving at a constant speed in the face of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written:

(7-13)

23 Summary of Chapter 7 • If the force is constant and parallel to the displacement, work is force times distance • If the force is not parallel to the displacement,

• The t ot al work i s th e work d one b y th e net force:

24 Summary of Chapter 7

• SI unit of work: the joule, J • TtlTotal work ki is equal lt to th e ch ange i n ki neti c energy:

where

Kinetic energy is either positive or 0, never negative.25 Summary of Chapter 7

• Work done by a spring force:

• Power is the rate at which work is done:

• SI unit of power: the watt, W

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