1. No category

Extra Problem of Chapter 5 to 8
Question 1
A box of textbooks of m = 24.8 kg rests on a loading ramp that makes an angle ! with
the horizontal. The coefficient of kinetic friction is µ k = 0.25 and the coefficient of static
friction is µ s = 0.35.
A. As the angle ! is increased, find the minimum angle at which the box starts to
slip.
n
+y
f
ax
n
f
ax
mg
θ
+x
θ
mg
y-component Using first law n = mg cos!
x-component If the box doesn’t then the friction force f depicted in the diagram is
a static friction force fs ! µ s n = µ s mg cos"
When the box begin to slip, the x-component of the weight (mg) equals the
maximum static friction force mg sin ! = µ s mg cos! .
cos!
1
; tan ! = µ s = 0.35 ; ! = 19.3!
=
sin ! µ s
B. At this angle, find the acceleration once the box has begun to move.
When ! = 19.3! x-component of the weight overcomes the static friction force and
the box accelerates down the incline. The friction force opposing the motion will be a
kinetic friction force f k = µ k n = µ s mg cos!
Using the above diagram and Newton’s second law:
mg sin " ! µ k mg cos" = ma x ;
(
)
(
)
a x = g sin ! " µ k g cos! : ax = 9.8m / s 2 sin19.3! ! 0.25 9.8m / s 2 cos19.3! ,
ax = 0.92m / s 2
C. At this angle, how fast will the box be moving after it has slid a distance 5.0 m
along the loading ramp? Use the kinetic equation
vx2 = v02x + 2ax x = 2ax x
(
)
vx = 2ax x = 2 0.92m / s 2 ( 5m ) = 3m / s
Question 2
In the diagram below block A has a mass of 4.00 kg and block B has mass 12.00 kg. The
coefficient of kinetic friction between block B and the table is µk = 0.25. Block B is
moving right and accelerating to the right with a = 2.00 m/s2.
a
v
The rope is
massless and
there is no
friction in the
pulley
B
A
C
a) Draw a free-body diagram of block A. Hence find the tension in the rope
connecting Blocks A and B.
y
Using Newton’s 2nd Law
T AB " m A g = m A a ! T AB = m A (g + a )
TAB
T AB = (4.00kg ) 9.8m / s 2 + 2.0m / s 2 = 47.2 N
(
mA g
)
a
b) Draw a free-body diagram of block B. Calculate the friction force on block B.
Determine the tension of the rope connecting blocks B and C.
y
Using Newton’s 1st Law for y-axis:
n " mB g = 0 ! n = mB g
Friction:
fk = nµ k = m B gµ k
n
x
a
TBC
ƒk
TAB
(
)
= (12.0kg ) 9.8m / s 2 ( 0.25 ) = 29.4NN
mBg
For x-component use Newton’s 2nd Law: TBC ! T AB ! f k = m B a
After some rearrangement
TBC = TAB + fk + mB a = 47.2N + 29.4N + (12.0kg ) 2.0m / s 2
(
)
TBC = 100.6 N = 101N
3C) Finally, draw a free-body diagram of block C. Determine the mass of block C.
y
Using Newton’s 2nd Law
TBC ! mC g = !mC a " TBC = mC (g ! a )
TBC
a
mC =
mCg
TBC
101N
=
= 12.9kg
g ! a 9.8m / s 2 ! 2.0m / s 2
Question 3
In the diagram below block A (100kg) is on a 30° incline is connected by a frictionless
pulley to block B (50 kg) on a 53.1° incline. (a) Which way will the system move when
released from rest. (b) What is the acceleration of the blocks? (c) What is the tension in
the cord?
mA = 100kg
mB = 50kg
30°
53.1°
FREE-BODY
DIAGRAM of B
FREE-BODY
DIAGRAM of A
+y
+y
a
+x
nA
nB
T
T
a
30° WA
+x
There is no
Friction
WB
53.1°
a) Since there is no friction, to find the acceleration, a, only the components of the forces
parallel to the incline (basically the x components). For Box A the x-component of the
weight is WAx = WA sin 30! = mA g sin 30! = (100kg ) 9.8m / s 2 ( 0.5 ) = 490N . For Box B
the x-component of the weight is
WBx = !WB sin 53.1! = !mB g sin 53.1! = ! ( 50kg ) 9.8m / s 2 ( 0.8 ) = !392N . Since the
(
)
(
)
magnitude of W Ax is greater than WBx , box A will accelerate down the incline and box B
up the incline, as shown in the diagram.
b) For Box A, using second Newton’s law on the x-component
Fxnet = W Ax ! T = m A a → 490 N ! T = (100kg )a .
[1]
For Box B, using second Newton’s law on the x-component
Fxnet = T ! W Bx = m B a → T ! 392 N = (50kg )a .
[2]
Adding [1] and [2] 98 N = (150kg )a → a = 0.65m / s 2 .
c) Substituting the answer for a into equation [1] we obtain
(
)
T = 490N ! (100kg ) a = 490N ! (100kg ) 0.65m / s 2 = 425N .
Alternatively, substitute the value of a into equation [2]
(
)
T = 392N + ( 50kg ) a = 392N ! ( 50kg ) 0.65m / s 2 = 425N .
Question 4
Block A of mass mA = 2.0 kg is moving up a 36.9° incline with a speed of v0 = 5.0 m/s. It
is attached to block B of mass mB = 4.0 kg, by a massless frictionless rope-pulley system.
The coefficient of kinetic friction between Block A and the surface of the incline is µk =
0.2. Block A moves up 0.5 m to the top of incline. Assume block A accelerates up incline
and B is accelerating downward as in diagram.
v0 = 5.0 m/s
a
BLOCK A
a
B
BLOCK B
36.9°
a) Draw a free-body diagram of block B, then use Newton’s second law to write an
equation that includes the tension of the rope, T and the acceleration, a.
+y
Newton’s 2nd law for y (vertical) component
Fynet = T ! mB g = !mB a
T ! 4.0kg " 9.8m / s 2 = ! ( 4kg ) a
T
T ! 39.2N = ! ( 4kg ) a ,
(1 point)
(1 point)
mBg
(e1)
a
b) Draw a free body diagram on block A showing all the forces on A and the
direction of its acceleration, a. Draw the x-y axes parallel and perpendicular to the
incline.
+y
a
+x
n
T
(2 points)
fk
θ
mA g
θ = 36.9o
c) Use the diagram in part b) to find the normal force (perpendicular to incline) and
the force of friction (parallel to incline) on block A.
y (vertical) component, use diagram of part b)
Fynet = n ! mA g cos 36.9! = 0 " n = 2.0kg # 9.8m / s 2 # 0.8 = 15.68N (1 point)
Kinetic friction fk = µ k n = 0.2 ! 15.68N = 3.14N (1 point)
d) Using diagram from part b, determine the x-component (parallel to incline) of the
net force, and use the second law to write an equation that include the tension of
the rope, T and the acceleration, a.
x (horizontal) component, use diagram of part b)
Fxnet = T ! fk ! mA g sin 36.9! = ma (1 point)
Use result of part c) Fxnet = T ! 3.14N ! 2.0kg " 9.8m / s 2 " 0.6 = ( 2.0kg ) a
T ! 14.9N = ( 2.0kg ) a , (e2) (1 point)
e) The two equations from part a) and d) are two equations with two unknown a and
T. Solve them to obtain the tension T and acceleration a. Partial answer
( a = 4.05m / s 2 ).
From part a), e1, T ! 39.2N = ! ( 4kg ) a " T = 39.2N ! ( 4kg ) a
From part d), e2, T ! 14.9N = ( 2.0kg ) a " T = 14.9N + ( 2kg ) a
Subs.
e2
into
e1
to
eliminate
T, 39.2N ! ( 4kg ) a = 14.9N + ( 2kg ) a " 24.3N = ( 6.0kg ) a
This give a = 4.05m / s 2 (1 point)
Subs. a = 4.05m / s 2 into e1 gives T = 39.2N ! ( 4kg ) 4.05m / s 2 = 23N (1 point)
(
)
Verify by Subs. a = 4.05m / s into e2 gives
T = 14.9N + ( 2kg ) 4.05m / s 2 = 23N
2
(
)
Question 5
In figure below small block of mass m rotates with radius r on a frictionless table at
constant speed v. It is attached by a massless rope to a box of mass M. Find the minimum
speed v such that the box M is stationary.b
v
+y
m
T
arad
T
r
+y
M
Mg
Examine Free-Body Diagram of mass M
Since it is stationary, use first law
T ! Mg = 0 → T = Mg , [1]
Examine Free-Body Diagram of mass m
Mass m is in uniform circular motion, use the second law
T = marad = m
v2
, [2]
r
Combining equations [1] and [2] Mg = m
v2
→v =
r
Mgr
m
QUESTION 6
In the diagram below a 30 kg crate is being moved at constant velocity by a force F
applied at 30° below horizontal. The coefficient of kinetic friction is µ k = 0.25 . A) Find
the magnitude of the applied force F. B) How much work is done by this force after the
crate has moved 4.5 m? C) How much work is done by friction for the same
displacement? D) How much work is done by the normal force n, and by gravity mg? E)
Find the net work done on the crate.
F
y
n
v
30°
x
30.0 kg
ƒk
mg
(A) y-component
(
)
F
F
+ ( 30kg ) 9.8m / s 2 = + 294N
2
2
!F
$
!F
$ F
Friction fk = µ k n = µ k # + 294N & = 0.25 # + 294N & = + 73.5N
"2
%
"2
% 8
F
x-component FxNET = F cos 30! ! fk = 0 " F cos 30! ! ! 73.5N = 0 " 0.741F = 73.5N
8
F = 99.2 N.
F
99.2N
Using the above fk = + 73.5N =
+ 73.5N = 85.9N .
8
8
(B) WA = Fx ( 4.5m ) = F cos 30! ( 4.5m ) = ( 99.2N ) ( 0.866 ) ( 4.5m ) = 387J . WA > 0 since
the x-component of the applied force is in the direction of displacement.
FyNET = n ! F sin 30! ! mg = 0 " n =
(C) W f = ! fk ( 4.5m ) = ! ( 85.9N ) ( 4.5m ) = !387J . Wf < 0 since the friction force is in the
opposite direction of displacement.
(D) The normal force and gravity is perpendicular to the direction of motion, so the work
done by these forces are zero.
(E) The total work done is WT = WA + W f = 387J ! 387J = 0. Note here that by the
work-energy theorem is W = !K = 0, and there is no change in kinetic energy of the box
(the speed of the box remains constant).
Question 7
In the diagram on the next page, a block A of mass mA = 2.0 kg is moving up a 36.9°
incline with a speed of v0 = 5.0 m/s. It is attached to block B of mass mB = 4.0 kg, by a
massless frictionless rope-pulley system. The coefficient of kinetic friction between
Block A and the surface of the incline is µk = 0.2. Using the method of energy, calculate
the speed of Block A after it has moved up 0.5 m to the top of incline.
position 2A
BLOCK A v0 = 5.0 m/s
position 1A
B
36.9°
B
position 1B
BLOCK B
position 1A
SOLUTION of 7 using Work-Energy Theorem
a) Draw a free-body diagram of block A, and determine the friction force on block
A, and the work done by this friction force after block A travels 0.5 m up
incline.
+y
+x
n
T
(1 point)
v
y-com Fynet = n ! mgcos 36.9 = 0
fk
mg
n = mgcos 36.9 = 2kg(9.8m /s2 )0.8 = 15.68N
36.9
friction fk = nµk = 15.68 N (0.2) = 3.136 N (1 point)
Work Done by friction
Wf = -(3.136 N) (0.5 m) = -1.57 J (1 point)
Wf < 0 since force of friction is opposite dispacement
b) Calculate the work done by gravity on block A after it has moved 0.5 m up
incline.
The x-component of gravity is -mg sin 36.9 = - 2kg (9.8 m/s2) 0.6 = - 11.76 N.
Work done on A. WgA = -11.76 N (0.5m) = -5.88 J. (1.5 point)
c) Calculate the work done by gravity on block B during same interval as part a)
and b).
Work done on B by gravity WgB = mg (0.5 m) = 4 kg (9.8 m/s2) (0.5 m) = 19.6 J.
This is positive since gravity on B is in the same direction as its displacement.
(1.5 point)
d) What is the work done by the tension in the rope on the system (block A + B)
during the same interval. HINT: You don’t need to do any calculation.
The work by the tension on A is T(0.5 m) since tension is in the same direction as
displacement of A.
The work by tension on B is -T(0.5m) since tension is opposite displacement of B.
The total work on system is T(0.5m) – T (0.5m) = 0. (1 point)
e) Using the work-energy theorem and the results of part a-d determine the speed
of Block A after it has moved 0.5 m up the incline.
Total work WT = Wf + WgA + WgB + WT = -1.57 J – 5.88 J + 19.6 J + 0 = 12.1 J (1
point)
Work-Energy Theorem WT = 9.8J = ΔK =
point)
with v1 = v0 = 5.0 m/s gives v 2 =
1
1
(mA + mB )v 22 ! ( mA + mB )v12 (1
2
2
! m $2
2(12.1J )
2(12.1J )
+ v12 =
+ #5 &
mA + mB
2.0kg + 4.0kg " s %
v1 = 5.39 m/s (1 point)
SOLUTION of 7 using Conservation of Energy
Equation 8-35 gives the conservation of total energy W = !Emec + !Eth + !Eint , where
W is the work done by on the system by an external force, Emec = U + K is the
mechanical energy that is the sum of the potential and kinetic energy, !Eth is the increase
in thermal energy due to friction (surface and air), and !Eint is the internal energy of the
system. The internal energy is a properties of the material (the box, the ball, the car …)
and for this course we set this to zero, !Eint = 0 . This gives W = !Emec + !Eth . For this
problem there is no external force, W = 0 , !Emec + !Eth = 0 .
The initial and final positions are shown in the diagram on the previous page.
The initial y positions: Box A position 1A ( y1A ), and Box B 1B ( y1B ).
The final y positions: Box A position 2A ( y2A ),, and Box B 2B ( y2B ).
For Box A: !U A = m A g ( y2A " y1A ) = m A g!yA . From the diagram
!yA = 0.5m sin 36.9! = 0.3m (> 0 since A rises), which gives
m
!U A = 2kg " 9.8 2 " 0.3 = 5.88J .
s
For Box B: !U B = m B g ( y2B " y1B ) = m B g!yB . From the diagram !yB = "0.5m (< 0
m
since B falls), which gives !U B = 4kg " 9.8 2 " #0.5m = #19.6J .
s
!U = 5.88J " 19.6J = "13.72J
For the kinetic energy !K = !K A + !K B = K A,2 " K B,1 + K B,1 " K A,1
m
, and final speed of both box A and B, v2
s
1
1
1
#1
& #1
& 1
!K = % m A v22 " m A v12 ( + % m B v22 " m B v12 ( = ( m A + m B ) v22 " ( m A + m B ) v12
$2
' $2
' 2
2
2
2
Initial speed of both box A and B, v1 = 5
!K =
2
1
1
m
( 6kg ) v22 " ( 6kg ) #%$ 5 &(' = ( 3kg ) v22 " 75J .
2
2
s
This gives !Emech = !U + !K = "13.72J + ( 3kg ) v22 " 75J = ( 3kg ) v22 " 88.72J .
From the last section the force of friction is fk = 3.136N and since block A moves 0.5m
up the incline, the increase in thermal energy is, !Eth = fk " 0.5m = 1.568J .
m
This gives !Emec + !Eth = 0 " ( 3kg ) v22 # 88.72J + 1.568J = 0 . This gives v2 = 5.39 ,
s
which is the same as obtained by the work-energy theorem
Question 8
Consider the system below
If the blocks is released from rest find the speed of the blocks after the 6.00 kg block has
descended 1.50 m. µ k = 0.250 . Use the Method of Energy.
This problem can be solved by combining Newton’s law and kinematic equations, but it
is easiest to use the work-energy method:
• Note that the work done by the tension T on the 8.00 kg block is positive and is
equal in magnitude to work done by the tension T on the 6.00 kg block, which is
negative. So the work done by the tension T in the rope is zero. Make sure you
understand this.
• Hence we need only consider the work done by the friction f k and gravitational
force, mg.
Work done by friction on 8 kg block is
! fk s = !mgµ k s = ! ( 8.00kg ) 9.8m / s 2 ( 0.250 ) (1.50m ) = !29.4J
Work done by gravitational force on 6.00 kg block is
mgs = ( 6.00kg ) 9.8m / s 2 (1.50m ) = 88.2J
The total work done on the system is
W = mgs ! f s s = 88.2 J ! 29.4 J = 58.8 J
Using the work-energy theorem and the fact that the initial speeds are zero
2W
1
vF =
W = !K = mv F2
m
2
Since the 8.00 kg and 6.00 kg blocks move together with the same speed, m is the
combined mass of the 6.00 kg and 8.00 kg block. Hence m = 14.00kg. Again make sure
2(58.8 J )
you understand this. So we obtain v F =
= 2.89m / s
14.00kg
(
(
)
)
Question 9
In figure below a 15 kg stone slides down a snow covered hill, leaving point A with a
speed of 10 m/s. There’s no friction between point A and B, but there’s friction after
point B, where it reaches the spring and compresses it till it comes to a stop. A) Find the
speed at the bottom of the hill (point B). B) Find the maximum compression of the
spring.
A
vA = 10 m/s
No friction
Rough surface with friction
µk = 0.2 and µs = 0.8
20 m
B
vB = ?
Spring at
equilibrium:
k = 2.00 N/m
15 m
As discussed in class for 1D motion, given an Fx vs. x graph the work done is the area
under the curve.Use conservation of energy from point A to point B, where only
gravitational potential energy is relevant: at point A, KA = ½(mvA2) and Ugrav,A = mgh; at
point B, KB = ½(mvB2) and Ugrav,B = 0.
1
1
U grav, A + K A = U grav, B + K B ! mvA2 + mgh = mvB2
2
2
vB = vA2 + 2gh =
(10m / s )2 + 2 ( 9.8m / s 2 ) ( 20m ) = 22.2m / s 2
(B)
vB = 19.8 m/s
20 m
fk = µkmg
B
15 m
vC = 0
C
100m
For point B to point C, !Emec + !Eth = 0 " !U el + !K + !Eth = 0
U el,B + K B = U el,C + K C + !Eth :
x
Fully compressed spring
!Eth = µ k mg(100 + x) is the work done by friction;
At point B, KB = ½(mvB2) and the elastic potential energy of the spring is Uel,B = 0;
At point C, KC = 0 and the elastic potential energy of the spring is Uel,B = ½(kx2).
1 2 1 2
1
1
mvB = kx + µ k mg (100m + x ) ! mvB2 " µ k mg (100m + x ) = kx 2
2
2
2
2
1
1
(15kg ) ( 22.2m / s )2 " ( 0.2 ) (15kg ) 9.8m / s 2 (100m + x ) = ( 2.00N / m ) x 2
2
2
2
3690J " #$ 2940J + ( 29.4N ) x %& = (1N / m ) x
(
)
x 2 + ( 29.4m ) x " 750m 2 = 0
Solution is
!29.4m ±
( 29.4m )2 + 4 ( 750m 2 )
!29.4m ± 62.2m
= 16.4m and -45.8m
2
2
The physical solution is x = 16.4 m and hence the maximum compression is 16.4 m.
x=
=