479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 484
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R1–7. The man A has a weight of 100 lb and jumps from
rest onto the platform P that has a weight of 60 lb. The
platform is mounted on a spring, which has a stiffness
k = 200 lb>ft. If the coefficient of restitution between the
man and the platform is e = 0.6, and the man holds himself
rigid during the motion, determine the required height h of
the jump if the maximum compression of the spring is 2 ft.
h
A
Conservation of Energy: The datum is set at the initial position of platform P. When
the man falls from a height of h above the datum, his initial gravitational potential
energy is 100h. Applying Eq. 14–21, we have
P
Conservation of Momentum:
T 1 + V 1 = T 2 + V 2
0 + 100h = 1 2 a 100
32.2 b(y M) 2 1 + 0
(y H ) 1 = 264.4h
m M (y M ) 1 + m P (y P ) 1 = m M (y M ) 2 + m P (y P ) 2
( +T) a 100
100
b(264.4h) + 0 = a
32.2 32.2 b(y M) 2 + a 60
32.2 b(y P) 2
[1]
Coefficient of Restitution:
e = (y p) 2 - (y M ) 2
(y M ) 1 - (y p ) 1
( +T) 0.6 = (y p) 2 - (y M ) 2
264.4h - 0
[2]
Solving Eqs. [1] and [2] yields
(y p ) 2 = 264.4h T (y M ) 2 = 0.4264.4h T
Conservation of Energy: The datum is set at the spring’s compressed position.
60
Initially, the spring has been compressed = 0.3 ft and the elastic potential
200
1
energy is
. Here, the compression of the spring caused by
2 (200) A0.32 B = 9.00 ft # lb
impact is (2 - 0.3) ft = 1.7 ft. When platform P is at a height of 1.7 ft above the
datum, its initial gravitational potential energy is
60(1.7) = 102 ft # lb . When
platform P stops momentary, the spring has been compressed to its maximum and
1
the elastic potential energy at this instant is
. Applying
2 (200)A22 B = 400 ft # lb
Eq. 14–21, we have
T 1 + V 1 = T 2 + V 2
1
2 a 60
32.2 b A 264.4hB2 + 102 + 9.00 = 400
h = 4.82 ft
Ans.
484