Problem 1 Solution:
(a)
N
T
F
F friction
T
T
m2
m2 g
m1
m1 g
(b) The block m1 moves in x-direction:
m1 a = F cos θ − T − N µ,
N = m1 g − F sin θ
The block m2 moves in y-direction:
m 2 a = T − m2 g
Put in the values of m1 , m2 , F , and µ, we obtain
1
12M g
4
)
3M a =4M g − T − (3M g −
5
3
5
M a =T − M g
Add the above two equation together
4M a = 2M g
We find a = 12 g.
For block m1 : a points to the right and |a| = a
For block m2 : a points up and |a| = a
(c) T = M a + M g = M g 32
1
T
Problem 2 Solution:
(a)
T1
y
T2
x
Mg
(b) Distance between the ball and the rod r =
= 6πL
.
Speed of the ball v = 2πr
T
T
√
52 − 42 L = 3L.
(c) The acceleration a points to the left.
2
2L
.
Its magnitude is |a| = vr = 12π
T2
(d) The the x- and y-components of the force on the ball from the upper string: F1 =
(−T1 53 , T1 45 )
The the x- and y-components of the force on the ball from the lower string: F1 = (−T2 35 , −T2 45 )
Newtons law for y-direction
4
4
0 = T1 − T 2 − M g
5
5
Newtons law for x-direction
3
3
−M |a| = −T1 − T2
5
5
We find
T1 − T2 =M g
5
4
T1 + T2 =M |a|
or
(e)
5
3
5
5
10π 2 L
5
T1 = M g + M |a| = M g + M
8
6
8
T2
5
5
10π 2 L
5
T2 = −M g + M |a| = −M g + M
8
6
8
T2
2
Problem 3 Solution:
(a)
N
T
T
M
M/2
g M/2
F spring
gM
(b) Assume the potential energy is zero when the string is unstretched. After we pull the
M block by a distance L, the potential energy of the M block is lower by M gL sin θ, the
potential energy of the M/2 block is increased by M2 gL, and the potential energy of the
spring is increased by 12 kL2 . So the total potential energy of the system is
U1 = −M gL sin θ +
1
M
1
gL + kL2 = kL2
2
2
2
When the M block moves to the position where the spring is unstretched, U1 is converted
to the kinetic energy K = 12 M v 2 + 12 M2 v 2 = U1 . We find
v=
U1
2
=L
M + M2
2k
3M
(c) Let L2 be the distance by which the M block moves up the plane from the position where
the spring is unstretched. The kinetic energy of the block M , 12 M v 2 , turns into its potential
energy L2 = M g sin θ:
1 2kL2
1
M gL sin θ = M v 2 = M
2
2
3M
We find
2kL2
kL2
,
or h = L2 sin θ =
L2 =
3M g
3M g
3
Problem 4 Solution:
(a) c
(b) a
(c) b
(d) e
(e) c
4