Problem 2
A punter accelerates a ball from rest to a speed
of 10 m/s during the time in which his toe is in contact with the ball
(about 0.20 s).If the football has
a mass 0.50 kg, what average force does the punter exert on the ball?
SOLUTION:
To find the average force we need to find the
average acceleration.
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KNOWNS
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UNKNOWN
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vo = 0
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a
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v = 10 m/s
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|
t = 0.20 s
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v = vo + at
a = (v – vo)/t = (10 m/s – 0)/(0.20 s)
a = 50 m/s2
F = ma = (0.50 kg)(50 m/s2) = 25 N
Problem 4
A
freight train has a mass of 1.5 x 107 kg.If
the locomotive can exert a constant pull of 7.5 x 105 N, how
long does it take to increase the speed of the train from rest to 80 km/h?
SOLUTION:
If
the locomotive can exert a net force of 7.5 x 105 N, then the
net acceleration is equal to this force divided by the mass of the train:
a
= F/m = (7.5 x 105 N)/( 1.5 x 107 kg) = .05 m/s2
Now
we solve the basic kinematics problem like we used to in previous chapters.
But first, convert 80 km/h into m/s :
80
km/h (1h/3600s)(1000m/km) = 22 m/s
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KNOWNS
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UNKNOWN
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vo = 0
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t
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v = 22 m/s
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a = 0.50 m/s2
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v = vo + at
t = (v – vo)/a = (22 m/s – 0)/(0.05 m/s2)
t = 440 s = 7.3 min
Problem 6
A 5.0-g bullet leaves
the muzzle of a rifle with a speed of 320 m/s.What
force (assumed constant) is exerted on the bullet while it is traveling
down the 0.82 m long barrel of the rifle?
SOLUTION:
To find the force, we
need to find the acceleration of the bullet as it traveled through the
barrel of the gun.
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KNOWNS
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UNKNOWNS
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|
vo = 0
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a
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v = 320 m/s
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|
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x – xo = 0.82 m
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v2 = (vo)2 + 2a(x – xo)
a = [(v2 – v2o)/2(x – xo)]
a = 62439 m/s2
F = ma = (0.005 kg)(62439 m/s2) = 312 N = 310 N
(sig. digits)
Problem 8
Two forces are applied
to a car in an effort to move it (see figure 4.8 in the book).
(a) What is the resultant
of the two forces?
SOLUTION:
F1
= 400 N at 300 to the +y-axis (or 600 from the x-axis)
F2
= 450 N at 1000 from the x-axis
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Force
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x-component
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y-component
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|
1
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400 cos 60 = 200
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400 sin 60 = 346
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2
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450 cos 100 = -78.1
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450 sin 100 = 443
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|
R
|
121.9
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789
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|R| = [(121.9)2 + (789)2] = 798 N
q = tan-1(789/121.9)
= 81.20 or 8.80 to the right of the y-axis.
(b) If the car has a mass of 3000-kg, what acceleration
does it have?
SOLUTION:
a = F/m = 798 N/3000 kg = 0.266 m/s2 = 0.27 m/s2
Problem 10
After falling from rest
at a height of 30 m, a 0.50-kg ball rebounds upward, reaching a height
of 20 m.If the contact between
the ball and ground lasted 2.0 ms, what average force was exerted on the
ball?
SOLUTION:
We must find the acceleration
of the ball during the time of contact.To
do this we must first find the velocity of the ball upon contact with the
floor:
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KNOWNS
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UNKNOWN
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vo = 0
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v
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y – y0 = -30 m
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|
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a = 9.80 m/s2
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v2 = v02 –2gy
v2 = -2(9.80 m/s2)(-30 m) = 588
v = 24.2 m/s
Then, we want to find the velocity of the ball
after it left the ground:
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KNOWNS
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UNKNOWN
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v = 0
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vo
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y – y0 = 20 m
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|
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a = 9.80 m/s2
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v2 = v02 –2gy
vo2 = v2 + 2gy = 0 + 2(9.80 m/s2)(20
m)
vo = 19.8 m/s
Now we can find the acceleration:
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KNOWNS
|
UNKNOWN
|
|
vo = 24.2 m/s
|
a
|
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v = 19.8 m/s
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|
|
t = 0.20 ms
|
|
v = vo + at
a = (v – vo)/t = (19.8 – 24.2)/(0.002 s) = -2200 m/s2
F = ma = (0.5 kg)(-2200 m/s2) = -1.1 x 103
N