Download Figure 1: Problem 1 Figure 2: Problem 2 1. The spring is unstretched

75g 600 m/s
4 kN/m
50 kg
7 kg
Figure 3: Problem 3
20
Figure 1: Problem 1
cylinders equals
Z
Z y2
m2 gdy +
Vg =
y20
y3
m3 gdy
y30
= m2 g(y2 − y20 ) + m3 g(y3 − y30 ) (3)
P
A
where m2 and m3 are the mass of the 2and 3-kg cylinders, y20 and y30 are the initial coordinates of the 2- and 3-kg cylinders. It has the relationship between (y2 −
y20 ) and (y3 − y30 ) as y2 − y20 = −2(y3 −
y30 ). We apply the work-energy equation to
2
the pulley system (1/2)m2 v02 + (1/2)m3 v30
+
m2 g(y2 − y20 ) + m3 g(y3 − y30 ) = (1/2)m2 v22 +
(1/2)m3 v32 . Substituting the constrained conditions into the work-energy equation gives
(1/2)m2 v02 + (1/2)m3 (−v0 /2)2 + m2 g(y2 −
y20 )−(1/2)m3 g(y2 −y20 ) = (1/2)m2 (−2v3 )2 +
(1/2)m3 v32 . Thus, y2 − y20 = 0.244 m
v0
2 kg
3 kg
Figure 2: Problem 2
1. The spring is unstretched when x = 0. If the
body moves from the initial position x1 = 100
mm to the final position x2 = 200 mm, (a)
determine the work done by the spring on the
body and (b) determine the work done on the
body by its weight.
Solution. The work done by the spring on the
body is
3. A 75-g projectile traveling at 600 m/s strikes
and becomes embedded in the 50-kg block,
which is initially stationary. Compute the enZ 0.2
2 0.2
1
ergy lost during the impact. Express your ankxdx = − (4000) x 0.1 = −60 J
U1−2 = −
swer as an absolute value | ∆E | and as a per2
0.1
(1)
centage n of the original system energy E.
The work done on the body by its weight is
Solution. Since the force of impact is internal to the system composed of the projectile
Z 0.2
and block and since there are no other extermg sin 20dx = 2.35 J (2)
U1−2 =
nal forces acting on the system in the hori0.1
zontal line of motion, it follows that the linear
2. Point P on the 2-kg cylinder has an initial vemomentum of the system is conserved.
locity v0 = 0.8 m/s as it passes position A.
m1 v1 i = (m1 + m2 )vi
(4)
Neglect the mass of the pulleys and cable and
determine the distance y of point P below A
where m1 is the mass of the projectile, m2 is
when the 3-kg cylinder has acquired an upward
the mass of the block, and v is the velocity afvelocity of 0.6 m/s.
ter
the impact. Thus v = 0.900 m/s The origSolution. The 2 and 3-kg cylinders are loinal
energy E is E = (1/2)(75)(10−3 )(6002 ) =
cated by the coordinates yA and yB mea13500
J The energy after the impact is E ′ =
sured from the fixed pulley. The total con−3
(1/2)[75(10
) + 50](0.9002 ) = 20 J. The enstant length of cable in the pulley system is
ergy
lost
is
∆E
= E − E ′ = 13480 J. n is
L = yA + 2yB + constants and 0 = v2 + 2v3
n
=
∆E/E
=
(13480/13500)(100)
= 99.9%
where L is constant. Differentiation with respect to time gives 0 = ẏA + 2ẏB . The work
done by the gravitational force acting on the
1