Transcript Problem 2 - Salisbury University

Physics 221
Chapter 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?
40 N
10 kg
Hawaii or
Bust!
600
Solution 5 . . . Lugging the Luggage
•
•
•
•
What is the speed when the distance is 3 m?
F.d = 1/2 m v2
40 N
(40 cos 600)(3) = (1/2)(10)(v2)
v = 3.5 m/s
• Moral of the story
• W = (F)(d)(cos)
10 kg
Hawaii or
Bust!
600
Scalar Product (Dot Product)
A . B = |A| |B| cos 
i.i =1
i.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.d
W = (3 i + 2 j) . 7 i
W = 21 + 0
W = 21
Salisbury
University
(b) | F| = 5 and |d| = 7 and F.d = 21
F.d = |F| |d| cos 
21 = (5)(7) cos 
cos  = 3/5
 = 53 0
Stretching Springs
Hooke’s Law: The amount of stretch is
directly proportional to the force applied.
F=kx
Lab Experiment
Slope = “k”
F
x
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=kx
mg = 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
W
W
W
W
=
=
=
=
=
F.d
(kx)(x)
kx2
(120)(0.3)2
10.8 J

Correct Solution 8 . . . Body building
• F is a variable force so integration must be
performed.
• W = F . dx
•
•
•
•
W
W
W
W
=  kx . dx
=1/2 kx2
= (1/2)(120)(0.3)2
= 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=
W=
W=
W=
F.d
mgh
2x10x3
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 nonconservative force. Example is friction. The total
mechanical energy (P.E. + K.E.) will not remain
constant in this case.
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