1 work & energy notes

Work & EnergyWork & Energy

In the past…In the past…

v, a, x, t v, a, x, t How things move, Kinematics How things move, Kinematics F, a, m F, a, m What makes them move, What makes them move, DynamicsDynamics

Now we will look at Now we will look at WHY WHY they move!!! they move!!! Energy!Energy!

Energy Energy The ability to do work. The ability to do work.

Work = Force x DisplacementWork = Force x Displacement

1 Joule = 1 Newton x 1 Meter1 Joule = 1 Newton x 1 Meter

1.1. Object must move.Object must move.

2.2. Force & Displacement must be on Force & Displacement must be on the same planethe same plane

3.3. Don’t forget air friction is negligible.Don’t forget air friction is negligible.

Examples of work done on a box:Work done by gravity? W = FG r = FG 0 = 0Work done by the table? W = FTB r = FTB 0 = 0

Can the table ever do work on the box?Work done by you? W = F cosθ r Work done by friction? W = FF r

How much work will the road do on an 1800 kg car when its brakes are applied, if the coefficient of friction between the road and the wheels is 0.5 and the car skids 6 m?

W = F r = Ff r = Fg r = m g r

= (0.5) (1800) (9.8) (6)W = 52, 920 J

F

r

F vs r Constant ForcesHow do you find work on

F vs r graph?Work = F r

= Area under the curve

F

d

F vs r

Dot Dot ProductProduct The The scalarscalar product product

of two vectorsof two vectors

AABB rememberemember that A r that A and B and B have to have to be in the be in the same same direction direction though.though.

AA

BB

AA

xx

AA

yy

AABB = = AA BB cos cos θθ

RulesRules•It’s commutative. A•B=B•AIt’s commutative. A•B=B•A•It’s distributive. It’s distributive. A•(B+C)=A•B+A•CA•(B+C)=A•B+A•C

Power = Work/TimePower = Work/TimePower = (Force times distance) / timePower = (Force times distance) / time

Power = Force (distance/time)Power = Force (distance/time)

Power = Force (velocity)Power = Force (velocity)

Tells you how much energy you use in Tells you how much energy you use in a certain amount of time.a certain amount of time.

Metric Unit: WattsMetric Unit: Watts

English Unit: Horsepower.English Unit: Horsepower.

746 Watts = 1 HP746 Watts = 1 HP

P = W/t or P = F vP = W/t or P = F v

Energy(J)

WorkW

Kinetic Energy

K

Potential Energy

U

Law of Conservation of Energy

EnergyEnergy

ability to do workability to do work

Kinetic Energy energy due to motion

K = ½ mv2

How much KE does a 1800 kg car going 25 mph (11.2 m/s) have?

K = ½ (1800)(11.2)2 = 113 000 JHow much work would friction need to do to stop it?

W = K = 113 000 J

Gravitational Potential Energy energy due to position

UG = mgh

Ex. You lift a 1.2 kg book from the first floor to your social studies class on the 2nd floor 5 m up. How much potential energy does the book have?

UG = m g h = (1.2) (9.8) (5) = 58.8 JHow much work did you do? Conservation!!! W = UG = 58.8 J

Law of Conservation of Energy Energy cannot be created or destroyedAll the energy in = All the energy outW + UG + KE = UG + KE (+ W) (Work out is done by friction. If no

friction, then no work out.)

A 600 kg roller coaster car is lifted to the top of A 600 kg roller coaster car is lifted to the top of the first hill, 55 m above the ground. the first hill, 55 m above the ground.

a. How much potential energy does it have?a. How much potential energy does it have?UUGG = m g h = m g h

b. How much work was done to get the cart to b. How much work was done to get the cart to the top of the first hill?the top of the first hill?

W = UW = UGG c. How fast is it going at the bottom of the first c. How fast is it going at the bottom of the first

hill?hill? U UGG = K = ½ mv = K = ½ mv22

d. If the second hill is 40 m high, then how fast d. If the second hill is 40 m high, then how fast will the cart be going when it crests the hill?will the cart be going when it crests the hill?

U UGG = U = UGG + K + K

In the first car example, was there any potential energy? No W + 0 + K1 = 0 + K2 (+0)

W = K2 – K1

W = ΔK

In the second example, was there any kinetic energy at the beginning or the end? No W + UUGG1 + 0 = UUGG2 + 0 (+0)

W = UUGG2 – UUGG1

W = ΔUGUG

Work Energy Theorem

W = ΔK W = ΔUUGG

In order to change any type of energy,

work must be done.

Sample ProblemSample ProblemA disgruntled physics student A disgruntled physics student drops her book off a 4 story drops her book off a 4 story building (12 m), how fast is the building (12 m), how fast is the book going before it hits the book going before it hits the ground?ground?

h = 12 mh = 12 mm = 1.7 kgm = 1.7 kg

Energy in = Energy outEnergy in = Energy out

UUGG + K = U + K = UGG + K + K

Double check with kinematics!Double check with kinematics!

h = 12 m

A school bus pulls into an intersection. A car traveling 35 km/h approaches and hits a patch of ice. The driver locks the brakes causing the car to slide toward the intersection. If the car is originally 26 m away and the coefficient of friction between the car’s tires and the icy road is 0.25, does the car hit the bus and poor innocent school children lose their lives … or does the car stop just in the nick of time letting the little children grow up to do physics problems involving school buses and icy roads?

Which of the following is an expression for mechanical power?

a. Ft/m

b. F2m/a

c. Fm2/t

d. Fm/t

e. F2t/m

A pendulum consisting of a mass m attached to a light string of length l is displaced from its rest position, making an angle θ with the vertical. It is then released and allowed to swing freely. Which of the following expressions represents the velocity of the mass when it reaches its lowest position?

a. (2gl(1-cos θ)) d. (2gl (1-sin θ))

b. (2gl tan θ) e. (2gl(cos θ-1))

c. (2gl cos θ)

A mass m is moving horizontally along a nearly frictionless floor with velocity v. the mass now encounters a part of the floor that has a coefficient of friction given by µ. The total distance traveled by the mass before it is slowed by friction to a stop is given by

a. 2v2/µgb. v2/2µgc. 2µgv2

d. µv2/2ge. µvg

A 0.75 kg sphere is dropped through a tall column of liquid. When the sphere has fallen a distance of 2.0 m, it is observed to have a velocity of 2.5 m/s.

a. How much work was done by the frictional “viscosity” of the liquid?

b. What is the average force of friction during the placement of 2.0 m?

*A rock of mass m is thrown horizontally off a building from a height h, as shown. The speed of the rock as it leaves the thrower’s hand at the edge of the building is vo. How much time does it take the rock to travel from the edge of the building to the ground?

a. (hvo) d. 2h/gb. h/vo e. (2h/g)

c. hvo/g

h

m vo

*What is the kinetic energy of the rock just before it hits the ground?

a. mgh b. ½ mvo

2

c. ½ mvo2 – mgh

d. ½ mvo2 + mgh

e. mgh – ½ mvo2

A 1.5 kg block is placed on an incline. A 1.5 kg block is placed on an incline. The mass is connected to a massless The mass is connected to a massless spring by means of a light string spring by means of a light string passed over a frictionless pulley, as passed over a frictionless pulley, as shown. The spring has a force shown. The spring has a force constant k equal to 100 N/m. The constant k equal to 100 N/m. The block moves down a distance of 16 block moves down a distance of 16 cm before coming to rest. What is cm before coming to rest. What is the coefficient of kinetic friction the coefficient of kinetic friction between the block and the surface of between the block and the surface of the incline?the incline?

k = 100 n/m

1.5 kg

35°