Download Solution of Midterm 2

MIDTERM #2, PHYS 1211 (INTRODUCTORY PHYSICS), 2014
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This exam book has 7 pages including an equation sheet on the last page
PART I: MULTIPLE CHOICE QUESTIONS (question 1 to 6)
For each question circle the correct answer (a,b,c,d or e).
1. (2.5 points) You are driving your car along the road and hit a patch that is so icy that
there is effectively no friction between your tires and the road. Which statement is most
accurate?
A) Your car will start to spin
B) Nothing you do will have any effect; without any friction, you will continue to move in a
straight line at a constant speed, as there is no net external force on you.
C) If you hit the accelerator, you will get out of the icy patch faster.
D) If you hit the brakes, you will eventually stop.
E) If you twist the steering wheel, you can turn into the skid.
ANSWER: B This was discussed in class!
2. (2.5 points) A 70-N block and a 35-N block are connected by a string as shown. If the
pulley is massless and the surface is frictionless, the magnitude of the acceleration of the
35-N block is:
Here is a quick solution:
Total mass
( 70N + 35N ) ÷ 9.8mis !2 = 10.71kg
Net force 35 N (70 N is cancelled by
normal force!)
a = F / m = 35N / 10.71kg = 3.26mis !2
A) 1.6 m/s2 B) 3.3 m/s2 C) 4.9 m/s2 D) 6.5 m/s2 E) 9.8 m/s2 ANSWER: B
3. (2.5 points) A boy pulls a wooden box along a rough horizontal floor at constant speed
by means of a force P as shown. In the diagram f is the magnitude of the force of friction,
N is the magnitude of the normal force, and Fg is the magnitude of the force of gravity.
Which of the following must be true?
A) P = f and N = Fg
B) P = f and N > Fg
C) P > f and N < Fg
D) P > f and N = Fg
E) None of these
At equilibrium so acceleration is zero
x-comp Fxnet = F cos! " f = 0 # F cos! = f # F > f
y-com Fynet = N + P sin ! " Fg = 0 # N = Fg " P sin ! # N < Fg
ANSWER: C
1
4. (2.5 points) Only if a force on a particle is conservative:
A) does it obey Newton's third law
B) does it obey Newton's second law
C) does it do no work when the particle moves exactly once around any closed path
D) it is not a frictional force
E) does the work it does equal the change in the kinetic energy of the particle
ANSWER C Read p. 179 to 181 on conservative forces.
5. (2.5 points) A particle moves 5 m in the positive x direction while being acted upon by a
!
"
"
"
constant force F = ( 4N ) i + ( 2N ) j ! 4Nk . The work done on the particle by this force is:
A) -20 J
B) 30J
C) 20J
D) 10 J
!
! E)! is impossible to calculate without
"knowing other forces
W = F • d = Fx d x + Fy d y + Fz dz , but d = 5mi , and W = 4N ! m = 20J
6. (2.5 points) A force of 10 N holds an ideal spring with a 20-N/m spring constant in
compression. The potential energy stored in the spring is:
A) 0.5J
B) 2.5J
C) 5J
D) 10J
E) 200J
"1
Hooke’s law Fx = kx ! x = 10N ÷ 20Nim = 0.5m .
1
1 N
2
ANSWER B
U spring = kx 2 = 20 ( 0.5m ) = 2.5J
2
2 m
PART II: FULL ANSWER QUESTIONS (question 7 to 10)
Do all four questions on the provided space. Show all works.
7. (10 points) In the diagram below block A ( mA = 2.0kg ) is on a frictionless surface.
Block B ( mB = 1.5kg ) rests on block A with the coefficient of friction (between block A
and B) being µ k = 0.33 (kinetic) and µ s = 0.45 (static). Block B is being pulled by a force
F =10N that causes an acceleration, a.
F = 10N
B
a
A
a) Assume for this part that block B does not slip and fall off block A. Draw a free body
diagram of all forces acting on the composite block A + B system. Hence determine the
acceleration, a, of block B (and A).
!
FN
(2 points)
A+B
a
F = 10N
F = ( mA + mB ) a ! a = 10N ÷ ( 3.5kg ) = 2.86mis "2
(1 point)
!
Fg, A+ B
2
7 (continued)
b) Draw a free-body diagram of all forces acting on block A. Assume that the system
accelerates with the acceleration of part a). Using the diagram calculate the direction and
magnitude of the force of friction acting on block A. Is this friction force static or
kinetic? Briefly explain your answer.
FN , ground ! normal force of ground on A
(2 points)
a = 2.86mis !2
fSAB ! static friction on A due to surface B
A
second law fsAB = mA a = 2kg ! 2.86mis "2 = 5.72N
Fg, A = mA g
(1 point)
Fg, B = mB g
Since the ground is frictionless, the only force that can accelerate block A is friction, and if we
assume that B do not slip on A, then the friction must be static. (1 point)
The normal force associated with the surface between A and B is the weight of B
Fg, B = mB g = 14.7N . Hence the maximum possible static friction is
fSAB, Max = Fg, B µ s = 14.7N ! 0.45 = 6.615N , which is greater than the actual static friction
fsAB = 5.72N . Hence there will be no slipping, and the friction is indeed static.
Finally note that the 10N force acts on B, and not on A.
c) Using Newton’s third law, calculate the magnitude and direction of the force of
friction acting on block B.
Since the surface of B exert a force to the right on block A of the magnitude fsAB = 5.72N .
Hence using Newton’s third law the surface of A must exert a force on block B to the left of
magnitude fsBA = fsAB = 5.72N (3 points)
!
8. (10 points) In the diagram below block A (mass 5.00 kg) moves up ( v ) a
! = 36.9! incline. It is connected to a hanging block B (mass unknown) by a massless
rope by a frictionless pulley system. The coefficient of friction between block A and the
incline is µ k = 0.25 and µ s = 0.4 .
!
v
A
B
θ
8 (continued)
a) Draw a free-body diagram of block A. The diagram should include all forces acting on
Block A as well as the direction of its acceleration. Use Newton’s second law to derive an
3
equation that includes the tension in the rope, T, and the acceleration, a. If the acceleration
m
of block A is 2.18 2 up the incline, calculate the tension in the rope.
s
+y
+x
a
FN
T
(1 point)
v
y-com Fynet = FN ! m A g cos 36.9 = 0
fk
mg
36.9
(
)
FN = mA g cos 36.9 = 5kg 9.8m / s 2 0.8 = 39.2N (1 point)
friction fk = nµk = 39.2 N (0.25) = 9.8 N (1 point)
x-comp Fxnet = T ! mg sin 36.9 ! fk = m A a (2 point)
m
m
T = m A g sin 36.9 + fk + m A a = 5kg ! 9.8 2 ! 0.6 + 9.8N + 5kg ! 2.18 2 = 50.1N (1 point)
s
s
b) Draw a free-body diagram of block B. The diagram should include all forces acting on
Block B as well as the direction of its acceleration. Use Newton’s second law to derive
another equation of the tension in the rope, T, and the acceleration, a. Using the result of part
a calculate the mass of block B.
T = 50.1N
(1 point)
B
a = 2.18
m
s2
Fg, B = mB g
Fynet = T ! mB g = !mB a " mB =
T
50.1N
=
= 6.57kg (3 points)
!2
g + a 9.8mis ! 2.18mis !2
9. (10 points) A block of mass 2.0 kg is sliding up an incline with a speed of 6.0 m/s. It
slides 2.0m up the incline, stops, after which it slides back down to its starting point.
When it reaches its initial starting point its speed is now 5.17 m/s. There is friction
between the block and the surface of the incline.
A) What is the change in kinetic energy ΔK during the stated interval?
2
2
1 2 1 2 1
m&
1
#
# m&
!K = mv2 " mv1 = ( 2.0kg ) % 5.17 ( " ( 2.0kg ) % 6 ( = "9.27J (3 points)
$
$ s'
2
2
2
s'
2
4
9 Continued
B) What is the work done by gravity on the block during the stated interval? Here you don’t
need to do any calculations, but you must justify your answer.
ZERO (1 point)
Let Wx be the x-component of gravity on block, which, of course, remains the same throughout
the motion. The work done as the block slides up is - Wx (2.0m) < 0 since gravity opposes
upward slide. For the slide down it is Wx (2.0m) > 0 since gravity aids sliding down. The total
work by gravity is Wg = - Wx (2.0m) + Wx (2.0m) = 0 (3 points)
C) Using the result of part A and B, and the work-energy theorem, calculate the work done
by friction on the block during the interval.
Hence the only work done is that by friction Wf, and using the work-energy theorem gives
W f = !K = "9.27J (3 points)
10. (10 points) Consider the system of two boxes below for which there is no friction (the
pulley does not rotate) and the pulley and rope are massless.
y = h = 5m, Ug = mgy
y0 = 0, Ug0 = 0
A) Assume that the system in the above diagram is at rest. Use the work-energy theorem or
conservation of mechanical energy to calculate the speed of the 4.00 kg box just before it hits
the ground.
Easiest to use Conservation of mechanical energy
Initial
Final
4 kg
2 kg
U g,initial
+ U g,initial
+ K initial = U g,4 kgfinal + U g,2 kgfinal + K final
,where y0 = 0, and Kinitial= 0
1
1
m4 kg gh + m2 kg gy 0 +0 = m4 kg gy0 + m2 kg gh + m2 kg v 2 + m4 kg v 2
2
2
v=
(
)
2 m4 kg ! m2 kg gh
m4 kg + m2 kg
=
2 ( 2kg ) " 9.8mis !2 " 5m
m
= 5.71
4kg + 2kg
s
5
(7 points)
B) In part A), calculate the change in potential energy of the 4.0 kg mass, and of the 2.0 kg
box.
For 4kg box, !U 4 kg = m4 kg gy0 " m4 kg gh = "m4 kg gh = "4kg # 9.8mis "2 # 5m = "196J
For 4kg box, !U 2 kg = m2 kg gh " m2 kg gy0 = m2 kg gh = 2kg # 9.8mis "2 # 5m = 98J
(3 points)
USEFUL EQUATIONS
!
"
"
"
A = Ax i + Ay j + Az k , unit vector notation.
! !
Scalar Product: Ai B = Ax Bx + Ay By + Az Bz = AB cos !
Solution of quadratic equation, ax 2 + bx + c = 0 ! x =
"b ± b 2 " 4ac
2a
1 2
ax t , vx = v0 x + ax t , vx2 = v02x + 2ax ( x ! x0 )
2
v2
g = 9.8m / s 2 , Fg = mg Centripetal acceleration arad =
r
! net
!
Newton’s Laws F = ! F = 0 (Object in equilibrium)
!
!
F net = ma (Nonzero net force)
Friction fs ! µ s FN , fk = µ k FN .
Work-Energy Theorem
! !
K = (1 / 2)mv 2 , W = F • d = ( F cos! ) d = F||d (Straight-Line Motion, Constant Force)
!
! !
" !
"
"
"
"
"
Scalar product form: F = Fx i + Fy j + Fz k , d = d x i + d y j + dz k , W = F • d = Fx d x + Fy d y + Fz dz
Kinematics x = x0 + v0 x t +
1 2 1 2
mv f " mvi (valid if W net is the net or total work done on the object)
2
2
! !
Hooke’s Law Fx = !kx . Work and Energy W = F • d = ( F cos! ) d = F||d ;
W net = !K =
W net = !K = (1 / 2)mv 2f " (1 / 2)mvi2 (valid if W net is the net or total work done on the
(
)
(
)
object); W grav = !mg y f ! yi (gravitational work), W el = ! (1 / 2)kx 2f ! (1 / 2)kxi2 (elastic work)
Conservation of Mechanical Energy (only conservative forces are present) Emech = U + K
W net = !"U = ! (U 2 ! U1 ) = "K = K 2 ! K1 , U1 + K1 = U 2 + K 2 , U grav = mgy , U el = (1 / 2)kx 2
Non-Conservative Forces Wexternal = !Emech + !Eth ( Wext work done by external forces, and we
set !Eint = 0 ), where !Eth = fk d (thermal energy or negative work done by friction).
(
) (
Using !Emech = !U + !K = U f " U i + K f " K i
)
Final energy U f + K f = U i + K i + Wext ! fk d initial energy plus external work
minus energy loss due to friction
6
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