Transcript Work and Power - Tenafly Public Schools

Work
AP Physics C
Mrs. Coyle
Work
• The work done on an object by an external
constant force:
W = F r cos q
• q is the angle between the displacement, r,
and the force F.
Products of two Vectors A and B
• Dot Product:
– A·B =A B cos q Scalar
• Cross Product:
– AxB = AB sin q
Vector
Work is a dot product
• Work is the dot product of Force and
displacement.
W= F·r
Units of Work
• Is a scalar quantity.
• Unit: Joule 1 J = 1 N∙m
• James Prescott Joule, English, 19thCentury
• Other units of work:
 British system:
 foot-pound
 cgs system:
 erg=dyne ·cm
 Sub-atomic level:
 electron-Volt (eV)
Meaning of Sign of Work
When the work done
by an external force
onto the system is:
+ : the system gains energy
- : the system loses energy
• When the force is in the direction of the
displacement, q=0o:
cos 0= 1 =>
W= F r and work is positive
Which force acting on the sled
does positive work?
When the force acts opposite to the direction of
the displacement, q=180o:
cos 180 o = -1 => W= F r and work is negative
Which force acting on the sled
does negative work?
When the force acts perpendicular to the
displacement,
then the work done by that force is zero.
What force(s) acting on the
sled does no work?
Question
• Does the centripetal force do work?
Example 1
• How much work was done by
the 30N applied tension force
to move the 6kg box along the
floor by 20m?
(A: 300J)
• How much work did the
weight do in the same case?
• If µ=0.2, how much work did
the friction do?
(Hint: Use Fxnet to calculate N)
(A: -136J)
Work done by Resultant
W= W1+ W2+ Wn
Wnet  Wby individual forces
Since work is a scalar, the net work done by
a resultant force is equal to the sum of the
individual works done by each individual
force acting on the object.
Work is the area under
an F vs d graph.
Work Done by a Variable Force
• For a small
displacement, Dx,
W ~ F Dx
• Over the total
displacement:
xf
W   Fx Dx
xi
Work Done by a Variable Force
xf
lim
Dx 0
 F Dx  
x
xf
xi
xi
Fx dx
xf
W   Fx dx
xi
• The work done is
equal to the area
under the curve.
Work
W = F Dr cos q
W =  F(x) dx
W =  F • dr
Example 2 – Work by a Variable
Force
The force acting on a particle is F(x)= (9x2 +3) N.
Find the work done by the force on the particle
as it moves it from x1 =1.5m to x 2=4m.
Ans: 199 J
Example 3 Work as dot product of
vectors
A particle is pushed by a constant force of
F=(4i +5j)N and undergoes a displacement of
Dr= (3i+ 2j)m.
Find the work done by the force.
Ans: (12 +10)J= 22J (Scalar Quantity)
Hooke’s Law
Force exerted by spring: Fs = - kx
–equilibrium position is x = 0
–k : spring constant (force constant)
(k increases as stiffness of the spring
increases)
Fs is a Restoring Force
• because it always acts to bring the object
towards equilibrium
• the minus sign in Hooke’s law indicates that
the Fs always acts opposite the displacement
form equilibrium.
Work Done by a Spring
Ws 

xf
xi
1 2
Fx dx  
kx  dx  kxmax

 xmax
2
0
Example #16
An archer pulls her bowstring back 0.4m by
exerting a force that increases uniformly from
0 to 230N.
a) What is the equivalent spring constant of the
bow?
b) How much work does the archer do in pulling
the bow?
Ans: a) 575N/m, b) 46.0J