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PHY 131
Ch 1 – 6 Exam A
Name
A ball is shot from the ground into the air. At a height of 10 m, its velocity is v = 8.0 i + 6.0 j.
(a) To what maximum height does the ball rise? What are the (b) magnitude and angle of the
ball's velocity just before it hits the ground?
ay = Δvy / Δt
so it will drop 11.8 m
|v| = (82 + 15.42)½
-10 = vf – v10m / Δt
y = ½ g t2
|v| = 17.4 m/s
2
-10 = 0 – 6 / Δt
11.8 = 5 tbottom
2/2
t10m to top = 0.6 sec 8/8
tbottom = 1.54 sec
8/8
vave =
Δy
/ Δt
½(6+0) = (ytop – 10m) / .6
ytop = 11.8 m
8/8
Or
y = ½gt2 + voy t
10 = -5t2 + [20/t – 6] t
t2 + 1.2 t = 2
t = 0.6 ± √2.36
t = 0.6 ± 1.54 sec
vave = Δy / Δt
½( v0+6) = 10/t
v0 = 20/t - 6
tan θ = -15.4 / 8
θ = 63 degrees below
horizontal
ay =
Δvy / Δt
10 = vbottom – vtop / Δt
10 = vbottom – 0 / 1.54
vbottom = 15.4 m/s 8/8
vx is constant = 8 m/s
Block B in the figure weighs 600 N. The coefficient
of static friction between block and table is 0.20;
angle θ is 30°; assume that the cord between B and
the knot is horizontal. Find the maximum weight of
block A for which the system will be stationary.
Ff – FTx = 0
FTx = Ff
12/12
FTx = μ FN
FTx = 0.2 (600)
FTx =120 N
FTy – FWA = 0
FTy = FWA 12/12
tan θ = Fy / Fx
9/9
FTy = FTx tan θ
FTy = 69.3 N
69.3 N = FWA
We see that
And
Ff – FTx = 0
FTy – FWA = 0
Why? ΣF = m a, if a = 0 then forces to the left =
right and forces up = down.
A roller-coaster car has a mass of 500 kg
when fully loaded with passengers. (a) if the
car has a speed of 20 m/s at bottom of the 10
m radius drop, what is the force exerted by
the track on the car at this point?
(a)
13/13
10/10
(b) What is the maximum
F
–
mg
=
m
v2 / r
N
speed the car can have at the
FN – 5000 = 500 202/10
top of the 15 m peak and still
FN = 25000 N
remain on the track?
(b )
10/10
FN + mg = mv2/r
Where FN goes to zero
mg = mv2/r
v = 12.2 m/s
PHY 131
Ch 1 – 6 Exam B
Name
At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 50 m
horizontally and h = 70 m vertically above the launch point. At the instant it reaches its
maximum height above ground level, what is its horizontal displacement D from the launch point?
y = ½ a t2 +vyt +0
70 = -5*22 +vy2 +0
vy = 45 m/s
10/10
x = ½ at2 +vxt +0
50 = 0 +vx2 +0
vx = 25 m/s
a = Δv
/ Δt
-10 = (vf – 45) / ttop
ttop = 4.5sec
8/8
8/8
vave = Δx / Δt
25 = Δx / 4.5
Δx = 112.5 meters
7/7
A student of weight 800 N rides a steadily rotating Ferris wheel (the
student sits upright). At the highest point, the magnitude of the normal
force on the student from the seat is 700 N. (a) What is the magnitude of
the normal force at the lowest point? (b) If the wheel's speed is doubled,
what is the magnitude FN at the highest point?
Top
(b)
13/13
2
FC = m(2v) /r = 400 N
mg – FN = 400 N
800 – FN = 400 N
FN = 400 N
Bottom
mg – FN = mv2/r
800 - 700 = mv2/r
FC = mv2/r = 100 N
10/10
FN – mg = mv2/r
FN – 800 = 100
FN = 900 N
10/10
A rocket of negligible mass moving in the horizontal
direction becomes attached a block pulley system.
Block A has mass of 3 kg, and Block B has a mass of 2
kg. The ramp is 30 degrees above the horizontal.
What thrust must the rocket exert to cause the block
system to accelerate up the ramp at a rate of 3 m/s2
if the frictional coefficient between block A and the
ramp is 0.2?
cos30 FThrust
5/5
FNet
= mT
a
– sin30mAg – mB g – Ff = (mA + mB ) a
5/5
5/5
14/14
5/5
cos30 FThrust – 15
– 20 – Ff
= (3 + 2) 3
cos30 FThrust –
μ FN = 15 +20 +15
cos30 FThrust – .2 (sin30 FThrust + cos30mAg) = 50 N
cos30 FThrust – .2 ( ½ FThrust +
26 ) = 50 N
FThrust = 72 N