ENPH 131 Assignment #8
Solutions
Problem 14.28
Roller coasters are designed so that riders will not experience a normal force that is more than 1.5 times their weight against
the seat of the car.
Determine the smallest radius of curvature Ρ of the track at its lowest point if the car has a speed of 25 ft/s at the crest of the
drop. Neglect friction.
The velocity of the car at the lowest point of the track can be determined using eqn 14-7, keeping in mind that the vertical
displacement, s = 120 - 10 = 110 ft, does positive work. W is the weight of the car.
1 W
2 32.2
T1 + ÚU1-2 = T2
1
H25L2 + WH110L = 2
W
32.2
v2
v2 = 7709 ft2 ‘ s2
The radius of curvature can be determined from the equation of motion in the normal direction, requiring N = 1.5W:
ÚFn = m an
W v2
32.2 Ρ
H7709L
H0.5L H32.2L
1.5 W - W =
Ρ=
Ρ = 479 ft
Problem 14.51
The material hoist and the load have a total mass of 900kg and the counterweight C has a mass of 250kg. At a given instant,
the hoist has an upward velocity of 4m/s and an acceleration of 1.4 m ‘ s2 .
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ENPH 131 assignment 8 solutions.nb
Determine the poswer generated by the motor M at this instant if it operates with an efficiency of Ε = 0.8.
Using the FBD’s below we can write the equations of motion for the hoist and the counterweight:
ÚFy h = mh ay
2 T + T ' - 900 H9.81L = 900 H1.4L
ÚFy C = mC ay
-T ' + 250 H9.81L = 250 H1.4L
From these two equations we find
T ' = 2102.5 N
T = 3993.5 N
Using eqn 14-10 we can find the power consumed:
Pout = F × v
Pout = 2 T × v
Pout = 2 H3993.3L H4L = 31 946.4 W
P
31 946.4
Pin = Εout = 0.8
Pin = 39 933 W = 39.9 kW
Pout = F × v
Pout = 2 T × v
Pout = 2 H3993.3L H4L = 31 946.4 W
P
ENPH 131 assignment 8 solutions.nb
3
31 946.4
Pin = Εout = 0.8
Pin = 39 933 W = 39.9 kW
Problem 14.79
Block A has a weight of 2.5lb and slides in the smooth horizontal slot.
If the block is drawn back to s = 2 ft and released from rest, determine its speed at the instant s = 0. Each of the two springs
has a stiffness of k = 150 lb  ft and an unstretched length of 0.5ft.
When the block is drawn back to s = 2 ft the extinction of the springs is
x2 =
22 + 22 - 0.5
x2 = 2.328 ft
At s = 0
x0 = 2 - 0.5
x0 = 1.5 ft
Now apply conservation of energy, eqn 14-21
T1 + V1 = T2 + V2
1
1 2.5
1
0 + 2A 2 H150L H2.328L2 E = 2 I 32.2 M v2 + 2A 2 H150L H1.5L2 E
v = 111 ft  s
Problem 14.90
The 0.4lb ball is shot from the spring device. The spring has a stiffness k = 11 lb  in and the four cords C and plate P keep the
spring compressed 3in when no load is on the plate. The plate is pushed back 3in from its initial position.
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ENPH 131 assignment 8 solutions.nb
If it is then released, determine the speed of the ball when it reaches a position 25in on the smooth inclined plane.
When the ball reaches its position, s = 25 in =
25
12
ft, it is
25
12
sin 30 °= 1.042ft above its original position and thus has gravita-
tional potential energy W h = (0.4)(1.042)ft lb. Now, apply conservation of energy of the final and initial positions. Note
k = 11 lb  in = 132 lb  ft:
0+
1
2
3+3 2
H132L I 12 M
ÚT1 + ÚV1 = ÚT2 + ÚV2
1 0.4
= 2 I 32.2 M v2 + H0.4L H1.042L +
v = 43.9 ft  s
1
2
3 2
H132L I 12 M
Problem 14.97
The 70kg man bungee jumps off the bridge at A with an initial downward speed of 1.5m/s.
Determine the required unstretched length of the elastic cord to which he is attached in order that he stops momentarily just
above the surface of the water. The stiffness of the elastic cord is k = 1 kN  m. Neglect the size of the man.
Conservation of energy at points A and B:
1
2
1
2
T1 + V1 = T2 + V2
1
m vA + m g hA = 2 kH150 - l0 L2
2
H70L H1.5L2 + 70 H9.81L H150L =
l0 = 136 m
1
2
H1000L H150 - l0 L2
ENPH 131 assignment 8 solutions.nb
1
2
1
2
T1 + V1 = T2 + V2
1
m vA 2 + m g hA = 2 kH150 - l0 L2
H70L H1.5L2 + 70 H9.81L H150L =
l0 = 136 m
1
2
H1000L H150 - l0 L2
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