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Problem 07-50
A 0.25 kg block is dropped on a relaxed spring that
has a spring constant of k = 250.0 N/m (2.5 N/cm).
The block becomes attached to the spring and
compresses it 0.12 m before momentarily stopping.
While being compressed,
A) What is the work done on it by gravity?
B) What is the work done on it by the spring force?
C) What is the speed of the block just prior to hitting the spring?
Problem 07-50
A 0.25 kg block is dropped on a relaxed spring that
has a spring constant of k = 250.0 N/m (2.5 N/cm).
The block becomes attached to the spring and
compresses it 0.12 m before momentarily stopping.
While being compressed,
A) What is the work done on it by gravity?
B) What is the work done on it by the spring force?
Problem 07-50
A 0.25 kg block is dropped on a relaxed spring that
has a spring constant of k = 250.0 N/m (2.5 N/cm).
The block becomes attached to the spring and
compresses it 0.12 m before momentarily stopping.
C) What is the speed of the block just prior to hitting the spring?
Problem 07-60
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0 ĵ + 7.0k̂ )
A) Work done on the particle
B) Average power
C) Angle between di and df
Problem 07-60
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0 ĵ + 7.0k̂ )
A) Work done on the particle
F is constant in this problem, so
(
v v v
W = F ⋅ df − di
)
= (3.0î + 7.0 ĵ + 7.0k̂ ) ⋅ ((−5.0 − 3.0)î + ((4.0 + 2.0) ĵ + (7.0 − 5.0)k̂ )
= (3.0î + 7.0 ĵ + 7.0k̂ ) ⋅ (−8.0î + 6.0 ĵ + 2.0k̂ )
= −24.0 + 42.0 + 14.0 = 32.0J
Problem 07-60
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0 ĵ + 7.0k̂ )
B) Average power
Pave = W/∆t = 32.0J/4s = 8.0W
Problem 07-60
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial
Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
v
d i = 38
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0 ĵ + 7.0k̂ )
v
d f = 90
C) Angle between di and df
di and df are two vectors, so take the dot product
v v
d f ⋅ d i = d f d i cos φ ⇒
cos φ = (−5.0î + 4.0 ĵ + 7.0k̂ ) ⋅ (3.0î − 2.0 ĵ + 5.0k̂ ) / ( d f d i )
(−15.0 − 8.0 + 35.0k̂ ) / ((9.49)(6.16) ) = 12.0 / 58.5 = 0.205
⇒ φ = cos −1 0.205 = 78.2o
Problem 07-70
m = 230kg, L = 12.0m
variable horizontal force F
horizontal distance d = 4.0m
A) magnitude of F in final position
B) Total Work done on crate
C) Gravitational Work
D) Work done by rope
E) Work done by force F
θ
Problem 07-70
m = 230kg, L = 12.0m
variable horizontal force F
horizontal distance d = 4.0m
θ
Problem 07-70
m = 230kg, L = 12.0m
variable horizontal force F
horizontal distance d = 4.0m
F = T sin θ = mg tan θ
θ
Daily Quiz, September 21, 2004
Which content do you prefer for the Tuesday lectures
without the scheduled exams?
1) Lecture over material as we’ve done so far.
2) Concentrate on working problems and examples.
3) No Lecture, just hold informal office hours.
Directions are Important in Space!
Vector (direction important) Quantities:
• displacement
• velocity
• acceleration
Scalar (direction not important) Quantities:
• distance
• speed
• acceleration (same word, but it really is a vector)
Components of Vectors
a x = a cos θ
a y = a sin θ
• Component notation vs
magnitude-angle notation
a=
a +a
tan θ =
2
x
ay
ax
2
y
Add vectors by components
r
a = a x î
+ a y ĵ
+ a z k̂
r
b = b x î
+ b y ĵ
+ b z k̂
r r r
r = a + b = (a x + b x )î + (a y + b y ) ĵ + (a z + b z )k̂
Scalar product
• Scalar product of two
vectors a and b
a.b = a b cos φ
a, b: magnitude of a, b
φ: angle between the
directions of a and b
r
a = a x î + a y ĵ + a z k̂
r
b = b x î + b y ĵ + b z k̂
r r
a ⋅ b = (a x b x ) + (a y b y ) + (a z b z )
Displacement
• Displacement is the
change in position (or
location)
∆x = x2 - x1
• Displacement is a vector
with both magnitude
and direction
Instantaneous velocity
• Velocity at a given instant
V = lim ∆t->0 (∆x/∆t) = dx /dt
– the slope of x(t) curve at time t
– the derivative of x(t) with
respect to t
– VECTOR! -- direction and
magnitude
• (Instantaneous) speed is the
magnitude of the
(instantaneous) velocity
Acceleration
• Acceleration is the change in velocity
• The average acceleration
v v
v
v 2 − v1 ∆v
v
a avg =
=
t 2 − t1
∆t
• (Instantaneous) acceleration
v
v
v
2v
d x
v dv
v dv d dx
⇒ a=
= ( )= 2
a=
dt
dt dt dt
dt
• Unit: m/s2
• It is correctly a vector!
Projectile Motion
Acceleration is constant and only acts in the −y direction
• The horizontal motion and the vertical motion are
independent of each other
• Horizontal motion: (Motion with constant velocity)
vx = v0x = v0cosθ0
x – x0 = ( v0cosθ0 ) t
• Vertical motion: (Motion of free-falling object)
vy = v0y + ayt = (v0 sinθ0 ) − g t
y − y0 = v0y t + ½ a t2 = (v0 sinθ0 ) t − ½ g t2
assume the upward direction is positive
Chapter 5 Force and Motion
• Force F
– is the interaction between objects
– is a vector
– causes acceleration
– Net force: vector sum of all the forces on an
object.
N v
v
v
v v v v
v
Ftotal ≡ Fnet = ∑ Fi = F1 + F2 + F3 + F4 + ... = ma
i =1
Newton’s Laws of Motion
Newton’s first law: If no force acts on a body, then the
body’s velocity cannot change, that is, the body can not
accelerate
–rest, still rest
–moving, continue moving with same velocity
Newton’s second law: The net force on a body is equal to
the product of the body’s mass and the acceleration of the
body:
ΣF=ma
Newton’s third law: When two bodies interact, the forces
on the bodies from each other are always equal in
magnitude and opposite in direction: FAB = − FBA
Newton’s Second Law
• Newton’s second law: The net force on a body is
equal to the product of the body’s mass and the
acceleration of the body
ΣF=ma
Σ F : vector sum of all the forces that act on that
body
Σ Fx = m ax
Σ F y = m ay
Σ F z = m az
ƒ Unit: 1 N = (1kg) .(1m/s2) = 1 kg.m/s2
Frictional Forces
A block lies on a horizontal floor.
a) What is the magnitude of the friction force (f) on it
f = 0N, but note fS,max = µSFN
from the floor?
b) If a horizontal force of 5 N is now applied to it,
but it does not move, what is f now? fs = 5N
c) If fs, max = 10 N, will the block move if the
horizontal applied force is 8 N? no, because F < fs
d) How about 12 N? yes, because F > fs
FN = mg
F
fS,max = µSFN
mg
Uniform circular motion
• Period of revolution (period)
T = 2πr/v
• Centripetal acceleration
magnitude: a = v2 /r
direction: radially inward
• v and a: constant magnitude
but vary continuously in direction
Centripetal force:
F = ma = mv2 /r (−r)
Work-Kinetic Energy Theorem
The change in the kinetic energy of a particle is
equal the net work done on the particle
rf
v v
1
1
2
2
∆K = K f − K i = mv f − mvi = Wnet = ∫ F ⋅ d r
2
2
ri
.... or in other words,
final kinetic energy = initial kinetic energy + net work
1
1
2
2
K f = mv f = K i + Wnet = mvi + Wnet
2
2
Work done by a variable force
∆Wj = Fj, avg ∆x
W = Σ∆Wj = ΣFj, avg ∆x = lim ΣFj, avg ∆x
∆x →0
xf
W = ∫ F( x )dx
xi
• Three dimensional analysis
xf
yf
zf
xi
yi
zi
W = ∫ Fx dx + ∫ Fy dy + ∫ Fz dz
=∫
rf
ri
v v
F ⋅d r
Power
• The time rate at which work is done by a force
• Average power
Pavg = W/∆t (energy per time)
• Instantaneous power
P = dW/dt = (Fcosφ dx)/dt = F v cosφ= F . v
Unit: watt
1 watt = 1 W = 1 J / s
1 horsepower = 1 hp = 550 ft lb/s = 746 W
• kilowatt-hour is a unit for energy or work:
1 kW h = 3.6 M J