Physics 221Chapter 7

Problem 1 . . . Work for slackers!

• WORK = Force x Distance

• W = F . D

• Units: Nm = J Newton meters = Joules

• Problem 1 : You push a car with a force of 200 N over a distance of 3 m. How much work did you do?

Solution 1 . . . Work

• W = F.d• W = 200x3• W = 600 J

200 N

Problem 2 . . . Energy

• Energy is the capacity to do work• Units : Joules (J)

• Kinetic Energy: Energy due to motion

• Problem 2 : What is the kinetic energy of the car in Problem 1?

Solution 2 . . . Kinetic Energy

Kinetic Energy = Work done (if no friction)K.E. = 600 J

• Problem 3 : Given that m = 1000 kg, what is the speed of the car?

Solution 3 . . . Speed and K.E.

• F = ma• 200 = 1000 a• a = 0.2 m/s2

• Vf 2= Vi

2 + 2ad

• Vf 2= 0 + 2(0.2)(3)

• Vf 2 = 1.2

• Vf = 1.1 m/s

Kinetic Energy : Formula

• K.E. = 1/2 m v2

• Problem 4 : The K.E. of a car is 600 J and its mass is 1000 kg. What is its speed?

Solution 4 . . . more K.E.

• K.E. = 1/2 m v2

• 600 = 1/2(1000)(v2)• v = 1.1 m/s

• Does this look familiar?

• Moral of the story: We can find the speed either by using Newton’s Laws or Energy Principles.

Problem 5 . . . Lugging the Luggage

• What is the speed when the distance is 3 m?

600

40 N

Hawaii or

Bust!

10 kg

Solution 5 . . . Lugging the Luggage

• What is the speed when the distance is 3 m?• F.d = 1/2 m v2

• (40 cos 600)(3) = (1/2)(10)(v2)• v = 3.5 m/s

• Moral of the story

• W = (F)(d)(cos)

600

40 N

Hawaii or

Bust!

10 kg

Scalar Product (Dot Product)

A . B = |A| |B| cos

i . i = 1i . j = 0

W = F . d

Problem 6 . . . work is a scalar !

F = 3 i + 2 j acts on an object and causes a displacement of 7 i

(a) How much work was done?(b) What is the angle between F and d?

Salisbury

University

Solution 6 . . . work is a scalar !

(a) W = F.dW = (3 i + 2 j) . 7 iW = 21 + 0W = 21

(b) | F| = 5 and |d| = 7 and F.d = 21

F.d = |F| |d| cos 21 = (5)(7) cos cos = 3/5

= 53 0

Salisbury

University

Stretching Springs

Hooke’s Law: The amount of stretch is directly proportional to the force applied.

F = k x

F

x

Slope = “k”Lab Experiment

Example 6.6 . . . Springy Spring

The spring constant (k) of a spring is 20 N/m. If you hang a 50 g mass, how much will it stretch?

Solution 6.6 . . . Springy Spring

F = k xmg = kx(50 /1000)(9.8) = (20)(x)x = 2.5 cm

Lab Experiment

Problem 8 . . . Body building

• How much work would you have to do to stretch a stiff spring 30 cm (k= 120 N/m)?

“Solution” 8 . . . Body building

• W = F . d• W = (kx)(x)• W = kx2

• W = (120)(0.3)2

• W = 10.8 J

Correct Solution 8 . . . Body building

• F is a variable force so integration must be performed.

• W = F . dx

• W = kx . dx• W =1/2 kx2

• W = (1/2)(120)(0.3)2

• W = 5 .4 J

Problem 1 . . . Potential Energy

• You lift a 2 kg book and put it on a shelf 3 meters high.• A. How much work did you do?• B. Was the work “lost”?

Solution 1 . . . Hidden Energy (P.E.)

• A. • W = F.d• W = mgh• W = 2x10x3• W = 60 J

• B.• Work is stored as Potential Energy (hidden).

Problem 2 . . . In other words

In other words, if the 2 kg book fell down from the top of the bookshelf (3m), what would its K.E. be?

Solution 2 . . . In other words

• The K.E. would be equal to the P.E. • K.E. = 600 J.

• In other words, energy was converted (transformed) from one form (P.E.) to another (K.E.)

Conservative Forces

• If the work done against a force does not depend on the path taken then that force is called a conservative force. Examples are gravity and spring force. The total mechanical energy (P.E. + K.E.) will remain constant in this case.

• If the work done against a force depends on the path taken then that force is called a non-conservative force. Example is friction. The total mechanical energy (P.E. + K.E.) will not remain constant in this case.

Vote Democrat . . . Just kidding!