Transcript Solution

Mechanics Exercise Class Ⅱ
Brief Review
Work and Kinetic Energy
 
W  Fd cos   F  d
Work Done by a Constant Force
(Work-Kinetic Energy Theorem)
K  K f  K i  W
Work Done by a Spring Force
1 2 1 2
Ws  kxi  kx f
2
2
Work Done by a Variable Force
Power
xf
yf
zf
xi
yi
zi
Ws   Fx dx   Fy dy   Fzdz
 
dW
P
 F v
dt
U  W    F  x dx
xf
Work and Potential Energy
xi
Gravitational Potential Energy U  mgy
U 
Elastic Potential Energy
Conservation of Mechanical Energy
The Center of Mass

1
rcom 
M

m
r
 ii
n
i 1
1 2 1 2
kx f  kxi
2
2
E mec  K  U
 



Macom  F1  F2  F3    Fn


Conservation of Linear Momentum Pi  Pf
1. The force shown in the Fig. has magnitude Fp=20N and
makes an angle of 300 to the ground .Calculate the work
done by this force when the wagon is dragged 100m along
the ground.
Solution:
We choose the x axis horizontal
to the right and the y axis
vertical upward. Then
Fp  Fx i  Fy j  Fp cos 300 i  Fp sin 300 j  17i  10 j
 
Whereas d=100m. Then using Eq W  F  d
 



W  F  d  17i  10 j   100i   17  100  1700 J
2. A 5.0kg block moves in a straight line on a horizontal
frictionless surface. Under the influence of a force that
varies with position as shown in Fig. 7-31 How much work
is done by the force as the block moves from the origin to
x=8.0m?
Solution: A block moves in a straight line, so
Eq7-36 can be simplified as
xf
W   Fx dx
xi
Key idea: the work done the system by
the force component Fx as the system
moves from xi to xf is the area under the
curve between xi and xf .
The work done is the area under the graph between x=0m
to x=8.0m is
1
1
W  ( 2  4 ) 10  ( 6  8 )  5  5  8  105 J
2
2
3. A block of mass 0.5 kg moves horizontally along a
frictionless rod attached to a cord of negligible mass. A
force T of constant magnitude 50N acts on one end of the
cord across a frictionless hoop. The block is moving
toward the right side at point A. What is the velocity of the
block when it is drawn to point B by force T?
Solution:
The work done by N、mg is zero.
T
A
From
B
1
1
m B 2  m A 2   T  dr
AB
2
2
4
4
4 x
0
0
(4  x)  3
WT   T  dr   T cos  dx   T
AB
Utilizing 

o A
mg
So
T
N
x
B
x
udu
2
u +a
 u +a
2
2
4
(4  x)
0
(4  x)  3
WT   50 
we get
2
2
2
d(4  x)
 50  (4  x)2  32 04  100(J)
m A2  2WT
B 
 20.9(m / s)
m
2
2
dx
4. (P157.44)Two identical coins are initially held at height h=11.0
m. Coin 1 is dropped at time t=0 and then lands on a muddy field
where it sticks. Coin 2 is dropped at t=0.500 s and then lands on
the field. What is the acceleration of the center of mass (com) of
the two-coin system (a) between t=0 and t=0.500 s, (b) between
t=0.500 s and time t1 when coin hits and sticks, and (c) between t1
and time t2 when coin 2 hits and sticks? What is the speed of the
center of mass when t is (d) 0.250 s, (e) 0.750 s, (f) 1.75s?
2
O
1
h
x
Solution:
Choose the dropping point as origin and downward
as positive direction, construct x coordinate.
t1 
2h
2  11m
2h


1
.
50
s
,
t

0
.
500
s

 2.00s
2
g
9.8m / s 2
g
(a) Between t=0 and t=0.500 s, coin 1 is free falling
and coin 2 is at rest, so

m g ˆ 1 ˆ
acom 
i  gi  4.9iˆ m / s 2
2m
2


(b) Between t=0.500 s and t1=1.50 s, coin 1 and 2 are both free
falling, and

m g  m g ˆ
acom 
i  giˆ  9.8iˆ m / s 2
2m


(c) Between t1=1.50 s and t2=2.00s, coin 1 is at rest and coin 2
is free falling, and

m g ˆ 1 ˆ
acom 
i  gi  4.9iˆ m / s 2
2m
2


(d) When t=0.250 s, coin 1 is free falling and coin 2 is at rest, and
the speed of com is
vcom 


mv1 1
1
  gt  9.8m / s 2 0.250s   1.225m / s
2m 2
2
(e) When t=0.750 s, 0.500s<t<t1, coin 1 and 2 are both free falling,
and
mv  mv2 1
1
vcom  1
 gt  g t  0.500 s   g 2t  0.500 s 
2m

2

1
9.8m / s 2 2  0.750 s  0.500 s 
2
 4.9m / s

2
(f) When t =1.75s, t1<t<t2, coin 1 is at rest and coin 2 is free
falling, and
vcom 


mv2 1
1
 g t  0.500 s   9.8m / s 2 1.75s  0.500s   6.125m / s
2m 2
2
5. (P154 21) In Fig.5-45, block 1 (mass 2.0 kg) is moving rightward
at 10 m/s and block 2 (mass 5.0 kg) is moving rightward at 3.0 m/s.
The surface is frictionless, and a spring with a spring constant of
1120 N/m is fixed to block 2. When the blocks collide, the
compression of the spring is maximum at the instant the blocks
have the same velocity. Find the maximum compression.
Solution:
The linear momentum of the two blocks and the spring system is
conserved, and the mass of the spring is negligible. Choose
rightward as positive direction and suppose when the compression
of the spring is maximum, the velocity of the blocks is v, we have
2.0kg10m / s  5.0kg3.0m / s  2.0kg  5.0kgv
yields v  5m / s
Because of the compression of the spring, the total kinetic energy
of the system decreased, gives us
1
1
1
1
2
2
2
2
kxmax  2.0kg 10m / s   5.0kg 3m / s    2.0kg  5.0kg 5m / s  
2
2
2
2

Substituting k=1120N/m into the above equation yields the
maximum compression of the spring:
xmax  0.25m