Download Solutions to Assignment #1

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Solutions to Assignment #1
FRICTION, Solutions (p. 71-2)
#4. What is the magnitude of the frictional force when pulling a 40.0 kg crate along a
level surface with a horizontal 300 N force if the coefficient of kinetic friction is 0.550?
ΣFn = 0 : Fnormal − mg = 0
Fnormal = mg = 40 × 9.81 = 392.4
since object is in motion :
Fkinetic = µ kinetic Fnormal = 0.55 × 392.4 = 215.8
The magnitude of the kinetic friction is 216 N.
#8. Compute the frictional force when a person tries to pull a 25.0 kg load up an incline
of 10.00 degrees if the pulling force of 150 N is applied at an angle of 60.0 degrees to the
angle of the incline and the coefficients of static and kinetic friction are 0.800 and 0.750,
respectively.
ΣFn = 0 :
Fnormal − mg cos 10 o + Fapplied sin60 o = 0
Fnormal = 25 × 9.81 × cos 10 o − 150 sin60 o =
= 241 .5 − 129 .9 = 111 .62
Fstatic = µ static Fnormal = 0.80 × 111 .62 = 89.30
Assume ΣFt = 0 :
Fequilbrium − mg sin 10 o + Fapplied cos60 o = 0
Fequilbrium = 25 × 9.81 × sin 10 o − 150 cos60 o
= 42.59 − 75.0 = −32.41
Since |Fequilibrium| is less than Fstatic friction is −32.4 N.
#12. Using figure 4.12 what applied force would be necessary to start the load moving
down the incline? (I.e., what applied force would be in opposite direction?)
ΣFn = 0 :
FBD
Fnormal − W cos10 = 0
o
n
+
t
Fapplied
Fnormal = 250cos10o = 246.2
Fstatic = µ staticFnormal = 0.50× 246.2 = 123.10 N
SinceΣFt = 0 and friction will be Fstatic :
Thus, Fstatic + Fapplied − W sin10 = 0
o
W = 250 N
θ = 10 E
µ static = 0.500
F friction
F normal
Fapplied = −123.10 + 43.4 = −79.7
Thus, the applied force will be 79.7 N directed down the incline (or –79.7 N).
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LINEAR KINEMATICS, Solutions (p. 87-8)
#2. A skater increases his velocity with constant acceleration from 22.0 m/s to 30.0 m/s
over a 4.00 second duration.
(a) What was his acceleration?
(b) What distance will be travelled during this period?
(c) What was his average velocity?
(d) What will be his instantaneous velocity after skating 30.0 metres?
(a)
v −v
30.0 − 22.0 8
= = 2.00
a= f i =
4.00
4
t
Acceleration is 2.00 m/s2.
(b)
1
s f = si + vi t + at 2
2
1
= 0 + 22(4) + (2.00)4 2 = 0 + 88 + 16 = 104.0
2
Distance will be 104.0 metres.
(c)
v = (vi + v f ) / 2 = (22 + 30) / 2 = 26.0
Average velocity is 26.0 m/s.
(d)
v 2f = vi2 + 2a ( s f − si )
2
2
v30
m = 22.0 + 2( 2.00)(30 − 0)
= 484 + 120 = 604
v30 m = ± 604 = ±24.58 m/s
Thus, the speed after running 30 metres is 24.6 m/s.
#4. A puck is shot along the ice with an initial velocity of 65.0 m/s and is decelerated at
the constant rate of 0.25 m/s2.
(a) How fast will the puck cross the goal line, 15.00 m away?
(b) How much time does the goalie have before the puck reaches her at the goall?
(a)
v 2f = vi2 + 2a ( s f − si )
2
v f = ± 65.0 2 + 2(−0.25)(15 − 0)
= ± 4217.5 = ±64.94
The puck will be traveling 64.9 m/s.
(b)
v f − vi 64.94 − 65.0
=
= 0.240 (or 0.400 if velocities have three sig. figs.!)
t=
a
− 0.25
Thus, the goalie will have 0.240 s to stop the puck. (Note you will get different answers
depending on how you rounded off the final velocity in part c.)
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#8. A golf ball is putted toward the cup that is 5.00 m away. If the deceleration of the
ball on the green is 0.250 m/s, at what velocity must the putt be made so that the ball just
drops into the cup? How much time will the ball take to reach the cup?
v 2f = vi2 + 2a ( s f − si )
2
Thus, vi = v 2f − 2a ( s f − si )
2
vi = 0 − 2(−0.25)(5 − 0) = 2.5
vi = ± 2.5 = ±1.5811
Thus, ball should start with a velocity of 1.581 m/s towards the cup.
v − vi 0 − 1.5811
t= f
=
= 6.325
a
− 0.25
It will take 6.33 seconds to reach the cup.
PROJECTILE MOTION, Solutions (p. 94-5)
#2. A high jumper takes off with a velocity of 2.50 m/s at an angle of 35.0o to the
horizontal. Her centre of gravity at takeoff was 1.200 m from the ground.
(a) What will be the maximum height of her centre of gravity?
(b) How far will she travel horizontally if she lands 1.000 metre above the ground
on a foam high jump pad?
(a)
First, find components of takeoff velocity:
vx = v cos θ = 2.5 cos 35° = 2.047 m/s
v y = v sin θ = 2.5 sin 35° = 1.4339 m/s
The components are (2.05, 1.434).
Then compute height given velocity is zero at top of flight using:
2
2
vfy = viy − 2g( sfy − siy )
sfy =
=
vfy2 −viy2
− 2g
+ siy
0 − ( 1.4339 )2
+ 1.200 = 0.10479 + 1.200 = 1.3048
− 2( 9.81 )
The high jumper will reach a height of 1.305 metres (130.5 cm) above ground.
(b)
First, calculate the velocity at the landing.
2
2
v fy = viy − 2 g (s fy − siy )
v 2fy = 1.43392 − 2(9.81)(1.000 − 1.200)
= 2.0561+ 3.924 = 5.980
v fy = ± 5.980 = ±2.445
The correct velocity is −2.445 m/s. Then calculate the time from takeoff to
landing.
t=
v fy − viy
−g
− 2.445 − 1.4339
t=
= 0.3954 s
− 9.81
Now calculate the horizontal distance traveled.
s x = vx t = 2.047(0.3954) = 0.8095 m
Thus, she will travel 0.810 m or 81.0 cm horizontally.
#4. A long jumper takes off with a velocity of 11.00 m/s at an angle of 25.0o. The
athlete’s centre of gravity was 1.250 m above the ground at takeoff and 0.500 m above
the ground at landing.
(a) Compute the rectangular components of the takeoff velocity.
(b) What will be the flight time?
(c) How far will the athlete travel horizontally during the flight phase?
(d) If the jumper increased the takeoff angle to 30.0o will he jump farther?
(a)
First, find components of takeoff velocity:
vx = v cos θ = 11.00 cos 25.0° = 9.969 m/s
v y = v sin θ = 11.00 sin 25.0° = 4.649 m/s
The components are (9.97, 4.65).
(b)
First, calculate the velocity at the landing.
2
2
v fy = viy − 2 g (s fy − siy )
v 2fy = 4.6492 − 2(9.81)(0.500 − 1.250)
= 21.61 + 14.715 = 36.33
v fy = ± 36.33 = ±6.027
The correct velocity is −6.027 m/s. Then calculate the time from takeoff to
landing.
v −v
t = fy iy
−g
− 6.027 − 4.649
= 1.0883 s
t=
− 9.81
The flight time is 1.088 s.
(c)
s x = vx t = 9.969(1.088) = 10.849
The jumper will travel 10.85 metres horizontally.
(d)
4
5
First, find components of takeoff velocity:
vx = v cos θ = 11.00 cos 30.0° = 9.526 m/s
v y = v sin θ = 11.00 sin 30.0° = 5.50 m/s
Next, calculate the velocity at the landing.
2
2
v fy = viy − 2 g (s fy − siy )
v 2fy = 5.502 − 2(9.81)(0.500 − 1.250)
= 30.25 + 14.715 = 44.97
v fy = ± 44.97 = ±6.706
The correct velocity is −6.706 m/s. Then calculate the time from takeoff to
landing.
v −v
t = fy iy
−g
− 6.027 − 5.50
= 1.244
t=
− 9.81
s x = vx t = 9.526(1.244) = 11.85
Therefore, yes the jumper will travel farther.