Seat:
PHYS 1500 (Spring 2007) Exam #2, V1
5 pts
1. As a pendululm swings back and forth, the forces acting on the suspended object are
the force of gravity, the tension in the supporting cord, and the air resistance. Which
of these forces does no work on the suspended object at all times during the motion?
(a)
(b)
(c)
(d)
5 pts
Name:
gravity
air resistance
tension in the supporting cord
all of them do some work
2. Give an example where the net torque on an object is 0, but the net force is not 0.
Explain how it works.
Many examples work; the requirement is that the object will accelerate but
not rotate. Possibilities...you are standing in an elevator that is accelerating
upwards, you are sitting in a car that is accelerating forward, a smooth ball
that is tossed in the air,...
5 pts
3. A 1700 N crate is being pushed across a level floor at a constant speed by a force F~ of
500 N at an angle of 37◦ below the horizontal. What is the coefficient of kinetic friction
between the floor and the crate?
The net force is 0. The x-component of the force is
−Ffric + F cos θ = 0
→
Ffric = F cos θ
The y-component of the force is
n − mg − F sin θ = 0
→
n = mg + F sin θ
The force from friction is related to the normal force by Ffric = µk n.
n = 1700 N + 500 N sin 37◦ = 2000 N
◦
n = 160 N + 100 N sin 53 = 240 N
500 N cos 37◦
= 1/5 = 0.2
2000 N
(V 1)
100 N cos 53◦
µk =
= 1/4 = 0.25
240 N
(V 2)
µk =
10 pts
4. A uniform ladder is 10 m long and has a mass of 4 kg. It rests against a vertical wall
and makes an angle of 53◦ with the floor. The coefficient of static friction between the
ladder and the floor is 0.5 and the coefficient of static friction between the ladder and
the wall is 0. Compute the force that the wall exerts on the ladder and the normal
and frictional force that the floor exerts on the ladder.
To solve this problem, you need to find the net force and the net torque
and make sure that each is 0. There are three forces acting on the ladder:
gravity, from the floor, from the wall. The force from gravity acts at the
center of the ladder. The wall only gives a force in the negative y-direction
which I call ny . The floor gives a normal force (I call nf ) in the y-direction
and the friction force (I call Ff ) in the x-direction. The x-component of the
net force gives
Ff + −nw = 0
→
Ff = n w
The y-component of the net force gives
nf + −mg = 0
→
nf = mg
The net torque must also be 0. To simplify this part, I will take the axis
of rotation to be where the ladder and the floor meet. Then there is only
torque from gravity and from the wall. The angle between the radial direction and gravity is 90◦ + θ. The angle between the force from the wall and
the radial direction is 180◦ − θ. Take the length of the ladder to be L. This
means the net torque gives
−(L/2) m g sin(90◦ + θ) + L nw sin(180◦ − θ) = 0
→
nw =
1 m g sin(90◦ + θ)
2 sin(180◦ − θ)
nf = 4 kg 10
m
= 40 N
s2
nw =
1 40 N sin 143◦
= 15 N
2 sin 127◦
Ff = nw = 15 N (V 1)
nf = 8 kg 10
m
= 80 N
s2
nw =
1 80 N sin 143◦
= 30 N
2 sin 127◦
Ff = nw = 30 N (V 2)
5 pts
5. If you press a book flat against a vertical wall with your hand, in what direction is the
friction force exerted by the wall on the book?
(a)
(b)
(c)
(d)
(e)
5 pts
downward
upward
out from the wall
into the wall
none of the above
6. The earth rotates once per day about its axis. Where on the earth’s surface should you
stand in order to have the smallest possible tangential speed? Explain.
The earth as a whole rotates with a fixed angular velocity. Therefore, you
need to be at the rotation axis to have the smallest tangential speed. This
would be at the north or south pole.
5 pts
7. A 40 kg pole vaulter running at 9 m/s vaults over the bar. Her speed when she is above
the bar is 3 m/s. Neglect air resistance, as well as any energy absorbed by the pole,
and determine her altitude as she crosses the bar.
Since the only force doing net work on the pole vaulter is gravity, we can
us KEi + P Ei = KEf + P Ef . The initial PE is 0 and the final PE is related
to the height that the pole vaulter reaches.
1
1
Mvf2 + Mgyf = Mvi2
2
2
1
yf =
2
1
yf =
2
"µ
m
9
s
"µ
m
8
s
¶2
¶2
µ
m
− 3
s
µ
m
− 2
s
→
¶2 #
¶2 #
yf =
´
1³ 2
vi − vf2 /g
2
/(10 m/s2 ) = 3.6 m
(V 1)
/(10 m/s2 ) = 3.0 m
(V 2)
10 pts
8. A 5 kg steel ball strikes a massive wall at 20 m/s at an angle of 53◦ with the plane
of the wall. It bounces off the wall with the same speed and angle. If the ball is in
contact with the wall for 0.2 s, what is the average force exerted by the wall on the
ball? What is the impulse delivered to the wall?
The average force exerted on the ball is F~ = ∆~p/∆t. You need to compute
this for both the x- and y-components. The impulse delivered to the wall is
minus the impulse delivered to the ball by Newton’s 3rd law. The impulse
is the average net force times the duration.
x-component:
pix = p sin θ = 5 kg 20 m/s sin 53◦ = 80 kg m/s
pfx = −p sin θ = −80 kg m/s
−160 kg m/s
= −800 N
(V 1)
0.2 s
pix = p sin θ = 4 kg 10 m/s sin 53◦ = 32 kg m/s
pfx = −p sin θ = −32 kg m/s
Fx =
Fx =
−64 kg m/s
= −320 N
0.2 s
(V 2)
y-component:
piy = p cos θ = 5 kg 20 m/s cos 53◦ = 60 kg m/s
0 kg m/s
=0N
0.2 s
piy = p cos θ = 4 kg 10 m/s cos 53◦ = 24 kg m/s
pfy = p cos θ = 60 kg m/s
Fy =
Fy =
(V 1)
pfy = p cos θ = 24 kg m/s
0 kg m/s
=0N
0.2 s
(V 2)
The impulse delivered to the wall is all in the x-direction and is
Ix = 800 N 0.2 s = 160 kg m/s
(V 1)
Ix = 320 N 0.2 s = 64 kg m/s
(V 2)
5 pts
9. Using a screwdriver, you try to remove a screw from a piece of furniture, but can’t get
it to turn. To increase the chances of success, you should use a screwdriver that
(a)
(b)
(c)
(d)
5 pts
has a narrower handle.
has a wider handle.
is longer.
is shorter.
10. Two masses, of size m and 3m, are at rest on a frictionless surface. A compressed,
massless spring between the masses is suddenly allowed to uncompress, pushing the
masses apart. What is the final speed of the light mass compared to that of the heavy
mass? (Be quantititave, for example ”1/5 as fast”.) Explain.
The light mass has a speed 3 times larger than that of the heavy mass. The
reason is that the net external force is 0 on this system. This means the
total momentum is conserved. The initial momentum is 0. This means the
ratio of the speeds must be the inverse of the ratio of masses.
5 pts
11. A car (mass 1000 kg) travels at a constant speed of 10 m/s on a level circular road with
a radius of 25 m. What is the angular acceleration? What is the magnitude of force
from friction?
The angular velocity is v/r. The angular velocity isn’t changing therefore
the angular acceleration is 0. The friction force is the only force that is
causing the car to move in a circle. Therefore, the friction force must equal
mv 2 /r.
Ff = 1000 kg (10 m/s)2 /25 m = 4000 N
(V 1)
Ff = 500 kg (10 m/s)2 /25 m = 2000 N
(V 2)
10 pts 12. A 20.0 kg block starts at the top of a greased ramp with a speed of 1 m/s; the ramp
is at an angle 53◦ above the horizontal. The block slides to the bottom; the total
distance travelled by the block is 5.0 m. (A) What is the block’s speed at the bottom
of the ramp if friction can be ignored? (B) Suppose the block’s speed at the bottom is
1.0 m/s less than that in (A). What is the force from friction?
For part (A), the only force that does work on the block is the force from
gravity. This means KEi + P Ei = KEf + P Ef with the final PE equal to 0.
This means
1
1
Mvi2 + Mgyi = Mvf2
→
vf2 = vi2 + 2gyi
2
2
The initial height can be found from the triangle yi = L sin θ.
µ
¶
m 2
+ 2 · 10
s
µ
¶
m 2
2
vf = 2
+ 2 · 10
s
vf2 = 1
m
(5 m sin 53◦ ) = 81
s2
m
(5 m sin 37◦ ) = 64
2
s
m2
s2
m2
s2
vf = 9 m/s
(V 1)
vf = 8 m/s
(V 2)
For part (B), the force from friction is the only force that gives a change
in mechanical energy: Wnc = ∆KE + ∆P E. Once you have the work from
friction, the force from friction can be found from W = −Ff d.
Wnc
1
= 20 kg
2
÷
m
8
s
¸2
·
m
− 1
s
¸!
−20 kg 10
Ff = −
Wnc
1
= 20 kg
2
÷
m
7
s
¸2
·
m
− 2
s
−170 J
= 34 N
5m
¸!
−20 kg 10
Ff = −
m
(5 m sin 53◦ ) = 630 J−800 J = −170 J
s2
(V 1)
m
(5 m sin 37◦ ) = 450 J−600 J = −150 J
2
s
−150 J
= 30 N
5m
(V 2)
Equations
Basic Mathematic Formulas
tan θ = ho /ha h2 = h2o + h2a
Acirc = πr2
4π 3
Circum of circ = 2πr
Vsph =
r
Asur of sph = 4πr2
3
cos 37◦ = 4/5 sin 37◦ = 3/5 cos 53◦ = 3/5 sin 53◦ = 4/5
sin θ = ho /h
cos θ = ha /h
Chapter 2
g = 10 m/s2
Avg speed = distance/elapsed time
∆x = xf − xi
v̄ = ∆x/∆t
1
∆x = (v0 + v)t
2
v = v0 + at
ā = (vf − vi )/(tf − ti ) = ∆v/∆t
1
∆x = v0 t + at2
2
v 2 = v02 + 2a∆x
Chapter 3
Ax = A cos θ
Ay = A sin θ
∆~r = ~rf − ~ri
A=
q
~vav = ∆~r/∆t
A2x + A2y
tan θ = Ay /Ax
~aav = (∆~v)/∆t
vx = v0x + ax t
1
∆x = (v0x + vx )t
2
1
∆x = v0x t + ax t2
2
2
vx2 = v0x
+ 2ax ∆x
vy = v0y + ay t
1
∆y = (v0y + vy )t
2
1
∆y = v0y t + ay t2
2
2
vy2 = v0y
+ 2ay ∆y
Chapter 4
X
~ = m~a
F
m1 m2
r2
~ 21
= −F
G = 6.67259 × 10−11
Fg = G
~ 12
F
f s ≤ µs n
N · m2
kg2
w = mg
f k = µk n
Chapter 5
W = (F cos θ)∆x
PE = mgy
1
KE = mv 2
2
Wnet = KEf − KE0
Wnc = KEf − KEi + PEf − PEi
P̄ =
Wg = mg(yi − yf )
W
∆t
Chapter 6
~p = m~v
~I = F∆t
~
~
F∆t
= ∆~p = m~vf − m~vi
m1~v1i + m2~v2i = m1~v1f + m2~v2f
P̄ = F v̄
Chapter 7
θf − θi
∆θ
=
tf − ti
∆t
1
∆θ = (ωi + ω)t
2
ωav =
ω = ωi + αt
vt = rω
at = rα
ωf − ωi
∆ω
=
tf − ti
∆t
1
∆θ = ωi t + αt2
ω 2 = ωi2 + 2α∆θ
2
αav =
v2
ac =
= rω 2
r
4π 2 3
T =
r
GMS
2
Chapter 8
τ = rF sin θ
I≡
X
mr2
Wnc = ∆KEt + ∆KEr + ∆P E
X
τ = Iα
L ≡ Iω
1
KEr = Iω 2
2
X
∆L
τ=
∆t