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Unformatted text preview: Problem 15.14 The force exerted on a car by at
prototype crash barrier as the barrier crushes is
F 2 7(4400 + 50,0003) N. where s is the distance in
meters from the initial contact. Suppose that you want
to design the barrier so that it can stop a 2400kg ear
traveling at l()() kmlh. What is the necessary effective
length of the barrier? That is, whttt is the dislanee
required for the barrier to bring the car to a step? Solution: A} l y I l); — ll
PM : mu*  —ml."l
L 2 f l] km
2 u“ e [00— : 27.78 E
h s i \‘f I
f — (4400 + 50.00U .\ MIX 7  'EJHI.“
ll 0 . if ]
7 {4400 .x' + 25.000 3") 1 : —;{2400)(27.78)2
t _ Solving A" L 6.00 m Problem 15.87 The bar is smooth. Use conservation
of energy to determine the minimum velocity the lO—kg
slider must have at A (a) to reach C; (b) to reach D. Solution: Choose A as the datum point. (:1) Suppose that the slider just reaches C; the velocity is zero and the
censervation of energy condition is émvﬁ : 2 mg. from which its : ‘/2g(2) : Zﬁ : 6.26 m/s . (b) To reach D, the slider must reach lhe top of the bar. Suppose that
it reaches the top of the bar with zero velocity. The conservation of energy condition is émvi : mg(2 + I). From which. 11,4, 2 2g(3) : ﬂ : 7.67 m/s Problem 15.20 In Problem 15.19, if the winch exerts
a tension T : T0(1 + 0.13) after the crate starts sliding,
what total work is done on the crate as it slides a distance
S : 3 m up the ramp, and what is the resulting velocity
of the crate? Solution: The work done on the crate is From the principle of work and energy, U : émuz. from which ,1 3 a
U :] Trix —f (In 'SinHM'x * m] (mgcostDu's. 2U
n n ‘1‘ u v : —ﬁ : 2.52 mls
m from which U : 1;. [(i + (ions1)]?1 — (mg sin 01(3) — Mung cosf})(3). From the solution to Problem [5.19, Ti. : 932.9 N411, from which the
total work done is U : 3218.4 — 1455.1  1253.9 : 509.36 Nim. Problem 15.26 Each box weighs 50 lb and the
inclined surfaces are smooth. The system is released
from rest. Determine the magnitude of the velocities of
the boxes when they have moved 1 ft. Solution: Write workwenergy for the system U z (50 lbsin45°)(l ft) — (50 lb sin 300)(l ft) 2 10.36 ftlb 1 100 lb
[0.36 flbl : 7 u2 :> y = 2.53 ms
2 32.2 ﬁfs Problem 15.63 The 4mkg collar is released from rest
at position I. Neglect friction. If the spring constant is x 6 kN/m and the spring is unstretched in position 2,
what is the velocity of the collar when it has fallen to
position 2'? 250 mitt L 200 min 4—i
Solution: Denote (l = 200 mm, It 2 250 mm. The stretch of tlte spring in position 1 is S. = «Nil + dz 7 d : U.l20 m and :112 53 = U.
The work done by the spring on the collar is t] (I l
Ummg :f (—kttdt : [—452] : 43.31 Nan.
(ll! 2 (H211 The work done by gravity is h
Ugmity z] (—mg)ds : mg}? : 9.8] Ntt‘t.
t) From the principle of work and energy Usm—ing + Ugva : %m 1:2, from which 2
[1: (_) (Umnng + Ugmily) : 5.l5 tit/s Problem 15.92 The spring constant k 2700 Nlm,
mA = [4 kg. and ms 2 l8 kg. The collar A slides on
the smooth horizontal bar. The system is released from
rest with the spring unstretched. Use conservation of
energy to determine the velocity of the collar A when it
has moved 0.2 m to the right. Solution: bet the dalum for b‘ be at its initial position, The stretch
ol'the spring when A has moved 0.2 m to the righl is $2 : ./(0.I5)2 + {0.3 +0.2)2 7 ./(0.15)2 + (0.3)? = 0.187111
T+V:T2+V21 0 + n : 514).:1 + also: + %(7t)0)(0t87)3 —(18)t9.81}t0.2). Solving. u : .20 mls ...
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 Spring '08
 N/A
 Dynamics

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