Download PROBLEM 13.3

PROBLEM 13.3
A baseball player hits a 5.1-oz baseball with an initial velocity of 140 ft/s at
an angle of 40° with the horizontal as shown. Determine (a) the kinetic
energy of the ball immediately after it is hit, (b) the kinetic energy of the ball
when it reaches its maximum height, (c) the maximum height above the
ground reached by the ball.
SOLUTION
Mass of baseball:
(a)
W
(5.1 oz)
m
W
g
1 lb
16 oz
0.31875 lb
32.2 ft/s 2
0.31875 lb
0.009899 lb s 2 /ft
Kinetic energy immediately after hit.
v v0 140 ft/s
T1
(b)
1 2
mv
2
1
(0.009899)(140)2
2
T1
97.0 ft lb
T2
56.9 ft lb
Kinetic energy at maximum height:
v v0 cos 40
1 2
mv
2
T2
Principle of work and energy:
Work of weight:
140cos 40
T1 U1
2
107.246 ft/s
1
(0.009899)(107.246)2
2
T2
U1
2
T2 T1
U1
2
Wd
40.082 ft lb
Maximum height above impact point.
d
(c)
T2 T1
W
40.082 ft lb
125.7 ft
0.31875 lb
125.7 ft
Maximum height above ground:
h 125.7 ft
2 ft
h 127.7 ft
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PROBLEM 13.9
A package is projected up a 15 incline at A with an initial
velocity of 8 m/s. Knowing that the coefficient of kinetic
friction between the package and the incline is 0.12, determine
(a) the maximum distance d that the package will move up the
incline, (b) the velocity of the package as it returns to its
original position.
SOLUTION
(a)
Up the plane from A to B:
TA
UA B
F
UA B
TA U A B
1 2 1W
W
mv A
(8 m/s)2 32
2
2 g
g
( W sin15 F )d
F
kN
0 N W cos15
W (sin15
TB : 32
W
g
0 N
0
0.12 N
W cos15
0.12cos15 )d
Wd (0.3747)
Wd (0.3743) 0
32
(9.81)(0.3747)
d
(b)
TB
d
8.70 m
Down the plane from B to A: (F reverses direction)
TA
UB
A
1W 2
vA
TB 0
2 g
(W sin15 F )d
W (sin15
UB
A
1.245W
TB U B
A
TA
d
8.71 m/s
0.12 cos15 )(8.70 m/s)
0 1.245W
1W 2
vA
2 g
v A2
(2)(9.81)(1.245)
vA
24.43
4.94 m/s
vA
4.94 m/s
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15
PROBLEM 13.17
The subway train shown is traveling at a speed of 30 mi/h when
the brakes are fully applied on the wheels of cars B and C,
causing them to slide on the track, but are not applied on the
wheels of car A. Knowing that the coefficient of kinetic friction
is 0.35 between the wheels and the track, determine (a) the
distance required to bring the train to a stop, (b) the force in
each coupling.
SOLUTION
k
0.35 FB
(0.35)(100 kips) 35 kips
FC
v1
(a)
30 mi/h
T1 U1
Entire train:
2
(0.35)(80 kips)
28 kips
v2
44 ft/s
0
T2
1 (80 kips 100 kips 80 kips)
(44 ft/s) 2
2
32.2 ft/s 2
(28 kips 35 kips) x
x 124.07 ft
(b)
0 T2
0
x 124.1 ft
Force in each coupling: Recall that x 124.07 ft
Car A: Assume FAB to be in tension
T1 V1
1 80 kips
(44) 2
2 32.2
2
FAB (124.07 ft)
T2
0
19.38 kips
FAB
FAB
T1 U1
Car C:
1 80 kips
(44) 2
2 32.2
( FBC
2
28 kips)(124.07 ft)
FBC
FBC
19.38 kips (tension)
T2
0
28 kips
8.62 kips
19.38 kips
FBC
8.62 kips (tension)
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