PHY121
Ch 2 – 6 Exam
On a hot summer day, a young girl
swings on a rope above the local
swimming hole. When she lets go
of the rope her initial velocity is
3.00 m/s at an angle of 30.0° above
the horizontal. If she is in flight
for 1.00 second, how high above
the water was she when she let go
of the rope?
y = ½ a t2 +
5/5
tan θ = ax / ay
tan θ = ac / g
tan θ = v2/r / g
tan θ = 33.3 / 10
θ = 73°
∑Fy = 0 thus
50 - ½ (20) –
40 –
Ff
= (5 + 2 + 3) a
=
10
a
4/4
8/8
8/8
Ff
8/8
3/3
40 – (µ2 FN2 + µ3 FN3) = 10 a
40–(.2*cos30°m2g + .3*m3g) = 10 a
40 – (
3.5 +
9 ) = 10 a
2
a = 2.75 m/s
14/14
20/20
10/10
2/2
Fup = Fdown
FTension-y = mg
cos θ = FTension-y / FTension
cos θ =
mg / FT
FT = mg / cos θ
1/1
Block 1, of mass = 5 kg , is connected over an
ideal pulley to block 2, of mass 2 kg and block
3, of mass 3 kg. How much time is required
for the system to travel 1 meter if μ2 = 0.2
and μ3 = 0.3? (either in terms of θ or let θ =30°)
=
mT
a
= (m1 + m2 + m3) a
vo-y t + yo
14/14
0 = ½(-10)12 + (sin30*3)1 + yo
0 =
-5
+
1.5
+ yo
yo = 3.5 m
At a popular ride at amusement
parks, people sit in a swing that
is suspended from a rotating
arm. Riders are at a distance
of 12 from the axis of rotation
and move with a speed of 20
m/s. What is theta?
Find Tension (in terms of m.)
FNet
m1g – sin30°m2g – Ff
Name:
3/3
x = ½ a t2
1 = ½ 2.75 t2
t = 0.85 sec
PHY121
Ch 2 – 6 Exam
x = ½at2 +
On a hot summer day, a young girl
swings on a rope above the local
swimming hole. When she lets go
of the rope 2 m above the water at
an angle of 60.0° above the
horizontal. If she travels 4
meters in the horizontal distance,
how much time was she in the air?
16/16
8/8
Block 1, of mass = 5 kg , is connected
over an ideal pulley to block 2, of
mass 3 kg and block 3, of mass 2 kg.
How much time is required for the
system to travel 1 meter if μ2 = 0.3
and μ3 = 0.2?
(either in terms of θ or let θ =30°)
See above for solution
vo-x
4/4
8/8
t + xo
4 = 0 + (cos60*v)t + 0
vo = 8/t
4/4 (attempt to solve)
y = ½ a t2 +
6/6
2
vo-y
8/8
t + yo
6/6
0 = ½(-10)t + (sin60*8/t)t + 2
0 =
-5 t2 +
6.93
+ 2
t = 1.34 sec
To give a 20 kg child a ride, two teenagers pull on a 4.0 kg sled with
ropes. Both teenagers pull with a force of 55 N at an angle of 35°
relative to the forward direction, which is the direction of motion.
In addition, the snow exerts a retarding force on the sled that
points opposite to the direction of motion, and has a magnitude of
57 N. Find a.
FNet
=
mT
a
(cos35°55 + cos35°55) – 57
33.1
a = 1.4 m/s2
Name:
6/6
= (msled + mchild) a
= ( 4 + 20 ) a