Multiple choice
Problem 1
A 50.0-N box is sliding on a rough horizontal floor, and the only horizontal force acting
on it is friction. You observe that at one instant the box is sliding to the right at 1.75 m/s
and that it stops in 2.25 s with uniform acceleration. The force that friction exerts on
this box is closest to:
a) 3.97 N b) 490 N c) 50.0 N d) 8.93 N
e) 38.9N
Final velocity (x) vx = 0
Mass of box is m = 50N ÷ 9.8m / s 2 = 5.1kg
vx = v0 x ! ax t, vx = 0
fk
v
a
0x = 1.75 m/s
x
a = v / t = 1.75m / s ÷ 2.25s = 0.778m / s
m
x
0x
2nd law taking left as positive
Fxnet = fk = max = ( 5.1kg ) .778m / s 2 = 3.97N
(
)
answer a)
Problem 2
A professor holds an eraser against a vertical chalkboard by pushing horizontally on it.
She pushes with a force that is much greater than is required to hold the eraser. The force
of friction exerted by the board on the eraser increases if she: WARNING: The correct
answer may surprise you. Think about the amount of static friction in each case.
A) pushes with slightly greater force B) pushes with slightly less force
C) pushes so her force is slightly downward but has the same magnitude
D) stops pushing E) pushes so her force is slightly downward but has the same magnitude
The three diagrams below shows the possible scenario. In all cases the normal force is
FN = Mg cos! ( ! = 0 for A and B). The maximum static friction is fsmax = FN µ s . We
will assume that in all cases, the upward (+ y, as indicated below) vertical force cancels
the downward forces so that the eraser do not fall.
A and B
C
E
fs
+y
fs
fs
Static Friction
+x Fcosθ
!
!
F
F
θ
Applied
Fsinθ
Fsinθ
θ
!
F
Force
Fcosθ
Mg
Mg
Mg
If the eraser is to remain on the wall (not fall) the y-component of the net force must
be zero: Fynet = 0 . The static friction fs < fsmax is calculated for all scenarios below.
A and B
y-component
Fynet = fs ! Mg = 0
fs = Mg
ANSWER: Clearly E!
C, y-component
+y (shown above)
Fynet = fs ! Mg + F sin " = 0
fs = Mg ! F sin "
E, y-component
+y (shown above)
Fynet = fs ! Mg ! F sin " = 0
fs = Mg + F sin !
Problem 3
A series of weights connected by massless cords are given an upward acceleration of 4
m/s2 by a pull P as shown below. A, B and C are the tensions in the connecting cords.
The smallest of the three tensions A,B, and C is closest to:
a. 483 N
pull P
b. 621 N
c. 196 N
d. 276 N
5.00 kg
e. 80.0 N
C
10.0 kg
B
15.0 kg
A
20.0kg
It is obvious that Tension A, TA, between 15.0kg and 20.0 kg mass, is the smallest
tension. To solve see free body diagram on 20.0 kg (m = 20.0kg) below:
ANSWER: D
a = 4.0 m/s2
Use 2nd law
Fynet = TA ! mg = ma
TA
(
TA = m(g + a) = 20kg 13.8m / s 2
TA = 276N
mg
)
Kinematics with Calculus
Problem 4 Kinematics with Calculus
(
) (
)
!
The position of a particle moving in an xy plane is given by r = 2t 4 ! 3 iˆ + t 5 ! 2t ˆj ,
!
with r in meters and t in seconds. A) find the average velocity and acceleration for the
time interval between t = 1s and t = 3s. B) find the velocity and acceleration at t = 1s in
unit-vector notation. C) What is the angle between the positive direction of the +x axis
!
and a line tangent (i.e. v ) to the particle's path at t = 1 s? Give your answer in the range
of (-180o; 180o).
!
A) r ( t ) = xiˆ + yˆj = 2t 4 ! 3 iˆ + t 5 ! 2t ˆj
(
) (
(
) (
!
at t = 3s, r ( 3s ) = ( 2 ( 3s ) ! 3) iˆ + (( 3s )
)
)
!
4
5
at t = 1s, r (1s ) = 2 (1s ) ! 3 iˆ + (1s ) ! 2 (1s ) ˆj = !1miˆ ! 1mˆj
4
5
)
! 2 ( 3s ) ˆj = 159miˆ + 237mˆj
!x ˆ !y ˆ x ( 3s ) " x (1s ) ˆ y ( 3s ) " y (1s ) ˆ
m
m
!
vavg =
i+
j=
i+
j = 80 iˆ + 119 ˆj
!t
!t
3s " 1s
3s " 1s
s
s
(
)
(
)
4
5
dx ˆ dy ˆ d 2t ! 3 ˆ d t ! 2t ˆ
!
ˆ
ˆ
v ( t ) = vx i + vy j = i +
j=
i+
j = 8t 3 iˆ + 5t 4 ! 2 ˆj
dt
dt
dt
dt
m
m
!
3
4
at t = 1s, v (1s ) = 8 (1s ) iˆ + 5 (1s ) ! 2 ˆj = 8 iˆ + 3 ˆj
s
s
m
m
!
3 ˆ
4
at t = 3s, v (1s ) = 8 ( 3s ) i + 5 ( 3s ) ! 2 ˆj = 216 iˆ + 403 ˆj
s
s
!vy ˆ vx ( 3s ) " vx (1s ) ˆ vy ( 3s ) " vy (1s ) ˆ
!v
m
m
!
aavg = x iˆ +
j=
i+
j = 104 2 iˆ + 200 2 ˆj
!t
!t
3s " 1s
3s " 1s
s
s
m
m
!
3
4
b) v (1s ) = 8 (1s ) iˆ + 5 (1s ) ! 2 ˆj = 8 iˆ + 3 ˆj
s
s
3
dvy ˆ d 8t ˆ d 5t 4 ! 2 ˆ
dv
!
a ( t ) = ax iˆ + ay ˆj = x iˆ +
j=
i+
j = 24t 2 iˆ + 20t 3 ˆj
dt
dt
dt
dt
mˆ
m
!
2 ˆ
3 ˆ
at t = 1s, a ( 3s ) = 24 (1s ) i + 20 (1s ) j = 24 2 i + 20 2 ˆj
s
s
m
m
!
3
4
C) v (1s ) = 8 (1s ) iˆ + 5 (1s ) ! 2 ˆj = 8 iˆ + 3 ˆj
s
s
3m
vx > 0 and vy > 0 ,1st quadrant ( !90 0 < " < 0! ). ! = tan "1
= 20.5! , ! = 20.5!
8m
(
(
(
) (
) (
) (
(
)
)
)
)
) (
( ) (
( )
(
)
(
) (
)
)
Work by Graphical Integration
Problem 5
The figure gives the acceleration of a 3.0 kg particle as an applied force moves it from
rest along an x axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set
m
by as = 5.0 2 .
s
HINT: For part a)
Use F = ma , and equation on
work in the back. For part b)
use the equation on power in
the back.
A) Calculate the work done after the particle move from xi = 0m to xf = 1.0m.
Repeat for xi = 0.5 m to xf = 1.0m. Repeat for the interval xi = 0 to xf = 5.0m. B) If
at x = 1m, the particle is at rest (at v = 0), calculate the speed of the particle when
it is at xf = 5.0m.
W=
xf
!x
i
xf
F dx = m ! a dx = m [ area under the graph ]
xi
For xi = 0 to xf = 1.0m:
1m
!0
a dx =
=
1
1
1
m
m2
( base ) ! ( height ) = (1m ) ( as ) = (1m ) "$# 5 2 %'& = 2.5 2
2
2
2
s
s
xf
W = m ! a dx = 3kg " 2.5
m2
= 7.5J
s2
For xi = 0.5m to xf = 1.0m: Here I will enlarge the figure for clarity
xi
as
1m
!0.5m a dx =
=
as/2
minus
0m
1m
0.5m
1m
0.5m
1.0m
1
1
!0.5m a dx = 2 ( base )( height )big " 2 ( base )( height )small
1m
=
W = m!
1
m
1
m
m2
(1m ) #%$ 5 2 &(' " ( 0.5m ) #%$ 2.5 2 &(' = 1.875 2
2
2
s
s
s
1.0m
0.5m
a dx = 3kg " 1.875
For xi = 0 to xf = 5.0m:
5m
!0
a dx =
=
+
= 5.625J
1
m
m
1
m
m2
(1m ) !#" 5 2 $&% + ( 3m ) !#" 5 2 $&% + (1m ) !#" 5 2 $&% = 20 2
2
2
s
s
s
s
xf
m2
s2
B) For xi = 1m to xf = 5.0m:
xi
!0
s2
+
W = m ! a dx = 3kg " 20
5m
m2
a dx =
= 60J
+
m2
! m$ 1
! m$
= ( 3m ) # 5 2 & + (1m ) # 5 2 & = 17.5 2
" s % 2
" s %
s
xf
m2
xi
s2
W = m ! a dx = 3kg " 17.5
= 52.5J
1 2 1 2
mv f " mvi , vi = 0 at x = 1m. The
2
2
1
2 " 52.5J
m
speed at x = 5m is W = 52.5J = mv 2f ! v f =
= 5.9
2
3kg
s
Use the work-energy theorem W = !K =
Angular Momentum
Problem 6
In diagram below a 2kg rock is at point P traveling horizontally with a speed of 12 m/s.
At this instant what is the magnitude and direction of the angular momentum? If the only
force acting on the rock is its weight, what is the rate of change (magnitude and direction)
of the angular momentum?
Direction perpendicular to x-y plane
! indicates +z out of the page
! indicates –z into page
Also
+x right
+y up
! ! !
! !
Angular Momentum L = r ! p = mr ! v valid for point particle w.r.t. point O
!
r , r = 8m
Using
rule on
! ! the! right! hand
!
L = r ! p = mr ! v it is easy to see that the
!
direction of L is –z or into the page
36.9°
143.1°
!
v
!
For the magnitude L = L = mvr sin143.1" = 2kg ! 12m / s ! 8m ! .6
L= 115.2kg-m2/s.
Torque !due to gravity on particle w.r.t. point O.
! !
! = r " Fg , Fg = mg = 2kg " 9.8m / s 2 = 19.6N
!
! ! !
r , r = 8m
Using the right hand rule on ! = r " Fg it is easy to see that the
!
direction of ! is +z or out the page.
53.1°
! = rFg sin 53.1! = 8m " 19.6N " .8 = 125.4Nim
36.9°
This can be expressed in a different unit
! = 125.4kg • m 2 / s 2
Using !second law for rotation in terms of angular momentum
! dL
!=
. Hence the net torque is the rate of change of angular momentum
!
dt
Fg
!
!
Since the rate of change of angular momentum ! = dL / dt has opposite direction (+z)
! ! !
! !
compared to the direction of the current angular momentum L = r ! p = mr ! v (-z), the
angular momentum is decreasing. Can you see the similarity with our much earlier
discussion on linear kinetics?
!
!
L
FINAL COMMENT AND ADVICE:
Angular
momentum
and
Torque
depend on
!
!
!
the origin (O). For example, L = 0 and ! = 0 may be zero for origin O, but nonzero
!
!
L ! 0 and ! " 0 in another origin O/. Study angular momentum problems and static
equilibrium problems.
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