PHYSICS 132
EXAM 3
STUDET’S FULL AME: ________________________________
ISTRUCTIOS:
- By placing your name on this exam and submitting it for credit, you are affirming this is your own
work and that you have neither given nor received unauthorized assistance in its completion.
1. How much work do movers do pushing (horizontally) a 160.0 kg crate 10.3 m across a rough floor
without acceleration, if the coefficient of friction is 0.50?
F = f k = µ k = µ k mg
for equilibrium
W = Fd cos θ = µ k mgd cos θ
W = (.5)(160kg )(9.8 sm2 )(10.3m) cos 0o
W = 8075 J
2.
At room temperature, an oxygen molecule, with mass of 5.31×10-26 kg, typically has a kinetic
energy of about 6.21×10-21 J. How fast is the molecule moving?
K = 12 mv 2
2K
(2)(6.21×10−21 J )
v=
=
m
5.31×10−26 kg
v = 484 ms
3.
An 88g arrow is fired from a bow whose string exerts an average force of 110N on the arrow over a
distance of 78cm. What is the speed of the arrow as it leaves the bow?
W = ∆K = K f − K i = 12 mv 2 − 0 = 12 mv 2
W = Fd cosθ
work related to change in K
definition of work
Fd cos θ = 12 mv 2
2 Fd cos θ
(2)(110 )(.78m) cos 00
=
v=
.088kg
m
v = 44.2 ms
4.
A spring has a spring constant, k, of 440N/m. How much must this spring be stretched to store 25 J
of potential energy?
1
2
k ∆x 2 = U
∆x 2 =
2U
(2)(25 J )
=
440 m
k
∆x = .34m
5.
Jane, looking for Tarzan, is running through the jungle at top speed, 5.3m/s, and grabs a vine
hanging vertically from a tall tree. How high can she swing upward?
Ei = E f
mechanical energy is conserved
Ki = U f
1
2
mv 2 = mgh
(5.3 ms ) 2
v2
=
h=
2g (2)(9.8 sm2 )
h = 1.43m
6.
A 145g baseball is dropped from a window 13m above the ground. If the ball hits the ground with a
speed of 8.00m/s, what is the average force of air resistance exerted on it?
∆E = WC = Fd cosθ
Work-Energy Theorem
E f − Ei = − Fd
F=
Ei − E f
d
F = 1.06 mgh − 12 mv 2 (.145kg )(9.8 sm2 )(13m) − 12 (.145kg )(8 ms )
=
=
d
13m
2
7.
What minimum power must a motor have to be able to drag a 310kg box along a level floor at a
speed of 1.20m/s of the coefficient of friction is 0.45?
F = f k = µ k = µ k mg
P = Fv
equilibrium
definition of instantaneous power
P = µ k mgv = (.45)(310kg )(9.8 sm2 )(1.2 ms )
P = 1640W
8.
What is the magnitude of the momentum of a 28.0g sparrow flying with a speed of 8.4m/s?
p = mv = (.028kg )(8.4 ms )
p = .24 kgs m
9.
A 23g bullet traveling 230m/s penetrates a 2.0kg block of wood and emerges cleanly at 170m/s. If
the block is stationary on a frictionless surface when hit, how fast does it move after the bullet
emerges?
Pi = Pf
conservation of momentum
mB vBi = mW vWf + mB vBf
vWf =
mB vBi − mB vBf
mW
vWf = 0.69 ms
(.023kg )(230 ms ) − (.023kg )(170 ms )
=
2kg
10. A ball of mass 0.440kg moving in the +x direction with a speed of 3.30m/s collides head-on with a
0.220kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of the
0.220kg ball after the collision?
m1 = .44kg
v1i = 3.3 ms
m2 = .22kg
v2i = 0
 2m1 
 m2 − m1 
 2m1 
+
=
v2 f = 
v
v
 1i 
 2i 
 v1i
m
+
m
m
+
m
m
+
m
 1
2 
 1
2 
 1
2 
 (2)(.44kg ) 
m
v2 f = 
 (3.3 s )
 .44kg + .22kg 
v2 f = 4.4 ms in + x direction
11. Block mA of mass 5.0 kg box is on a horizontal surface which has a coefficient of kinetic friction
0.45. It is connected to a hanging block mB of mass 8.0 kg by a light string over a massless,
frictionless pulley so that mB hangs at the end of the string. When released from rest, what will be
the acceleration of mA? (Draw complete force diagrams and show each force in component form.)
Forces on mA :
r
0iˆ
=
r
T=
Tiˆ
r
0iˆ
WA =
r
f k = − µ k iˆ
mA
mB
∑F
∑F
∑F
+
+
+ −mAgˆj
+
0 ˆj
y
= − mAg = 0 ⇒ = mAg
x
= T − µ k = T − µ k mAg = mA a
x
= mB g − T = mB a ⇒ T = mB g − mB a
(mB g − mB a) − µ k mAg = mA a
jˆ
0 ˆj
substitute into above eq.
solve for a
mB g − µ k mAg (8kg )(9.8 sm2 ) − (.45)(5kg )(9.8 sm2 )
a=
=
8kg + 5kg
mA + mB
a = 4.33 sm2
Forces on mB :
r
T=
−Tiˆ + 0 ˆj
r
WB = mB giˆ + 0 ˆj
Extra Credit
A box of mass 2.5kg is to slide down a frictionless ramp to
a vertical loop of radius 2.0m. If the box is to just make it
through the loop without falling off at the top, what
minimum height, h, must the box be released from?
At Point B the only force on the box is its weight, the
normal is 0. Since the box is in circular motion, the net
force on it is centripetal.
At B:
mvB2
FET = mg =
r
vB = r g
By conservation of mechanical energy:
E A = EB
mghA = 12 mvB2 + mghB
hA = h
hB = 2r
m cancels
vB2 = rg
gh = 12 (rg) + g(2r ) = 2.5rg
h = 2.5r = (2.5)(2m)
h = 5m