No Title
PHYSICS I
Solution to Exam 3
                                                                                                           
Monday 10/26/15 Name
O101: periods 1,2
JFW CM:
Section:














                    
                                         (1)
(25)
(2)
(25)
(3)
(25)
(4)
(25)
TOTAL
(100)
  1. A mass m travels with kinetic energy K along a smooth horizontal surface. The mass encounters an ideal spring of constant k compressing it and then rebounds.
    m=2.40 kg
    K=0.588 J
    k=60.0 N/m

      (a) [5 pts] With what speed does the mass approach the spring?

      K
      =
      1
      2
      		 mv2
      0.588 J
      =
      1
      2
      		 (2.40 kg)v2
      v
      =
      0.700 m/s

      (b) [7 pts] Calculate the total impulse which the spring exerts on the mass while contact is maintained (assume no energy loss).

      Jsp,m
      =
      m(vf-vi)
      =
      (2.40 kg) �
      �       		(0.700 m/s 
      ^
      i
       
      		 )-(-0.700 m/s 
      ^
      i
       
      		 ) �
      �
      =
      3.36 N·s 
      ^
      i
       
      where, the positive x-axis was chosen in the direction of the final velocity.
      (c) [7 pts] If the mass contacts the spring for 0.628 s, what average force did the spring exert on the mass.

      Fav
      =
      J
      Dt
      =
      3.36 N·s 
      ^
      i
       
      0.628 s
      =
      5.35 N 
      ^
      i
       

      (d) [6 pts] By how much was the spring compressed?

      U
      =
      K
      =
      1
      2
      		 k(Dx)2
      0.588 J
      =
      1
      2
      		 (60.0 N/m)(Dx)2
      Dx
      =
      0.140 m
  2. A block of mass m is pulled up a rough ramp by an applied force F which is parallel to the ramp. The ramp angle is q. The frictional force opposing the motion is f. The magnitudes are

    m=3.40 kg
    F=15.0 N
    f=1.20 N
    q = 20.0o

      (a) [4 pts] Draw a free body diagram for the mass.
      ANS: There are four forces acting on the block. Take the positive x-axis up the ramp. The normal force is perpendicular to the ramp pointing in the positive y-direction. The applied force is parallel to the ramp in the positive x-direction. The frictional force is parallel to the ramp in the negative x-direction. The gravitational force is straight down.
      (b) [4 pts] State whether each force does positive or negative work.
      ANS: The normal force does NO work. The applied force does POSITIVE work. The frictional force does NEGATIVE work. The component of gravity parallel to the ramp does NEGATIVE work. The component of gravity perpendicular to the ramp does NO work.
      (c) [10 pts] Calculate the acceleration of the block up the ramp.

      �
      		Fx
      =
      F-f-mgsinq
      =
      max
      ax
      =
      F-f
      m
      		 -gsinq
      =
      15.0 N-1.20 N
      3.40 kg
      		 -(9.80 m/s2)sin20.0o
      =
      0.707 m/s2

      (d) [7 pts] If the mass was at rest initially, what speed will the block acquire after traveling 2.00 m up the ramp?

      v2f
      =
      v2i+2axDx
      =
      0+2(0.707 m/s2)(2.00 m)
      =
      2.83 m2/s2
      vf
      =
      1.68 m/s
  3. A block of mass m rests on a horizontal floor. An unknown constant, vertical force F=+F [^(j)] is suddenly applied to the block causing it to rise a height h while reaching a speed v.
    m=6.00 kg
    F=?
    h=8.00 m
    v=10.0 m/s

      (a) [10 pts] Calculate the total work done by all forces acting on the block.
      Use the work-energy theorem.
      Wtot
      =
      DK
      =
      1
      2
      		 mv2f- 1
      2
      		 mv2i
      =
      1
      2
      		 (6.00 kg)(10.0 m/s)2
      =
      300 J

      (b) [5 pts] Calculate the net force acting on the block.

      Fnet
      =
      Wtot
      Dy
      =
      300 J
      8.00 m
      =
      37.5 N

      (c) [5 pts] Calculate the acceleration of the block from information already determined.

      ay
      =
      Fnet
      m
      =
      37.5 N
      6.00 kg
      =
      6.25 m/s2

      (d) [5 pts] Calculate the magnitude of the unknown force, F. You can use dynamics here.

      �
      		Fy
      =
      F-mg
      =
      may
      F
      =
      m(g+a)
      =
      (6.00 kg)(9.80 N/kg+6.25 N/kg)
      =
      96.3 N
  4. The momentum of a particle tarveling along the x-axis is given by
    px(t)
    =
    at-bt2
    where,
    a=1.40 kg·m/s2
    b=2.80 kg·m/s3

      (a) [8 pts] Determine the equation describing the instantaneous force acting on the particle.

      Fx(t)
      =
      dpx(t)
      dt
      =
      a-2bt

      (b) [5 pts] At what time will the acceleration of the particle be zero?
      The acceleration of the particle will be zero wneh the net force is zero.
      a-2bto
      =
      0
      to
      =
      a
      2b
      =
      1.40 kg·m/s2
      2(2.80 kg·m/s3)
      =
      0.250 s

      (c) [8 pts] Calculate the average force acting on the particle in the time interval
      1.00 s � t � 2.00 s

      Fx-av
      =
      px(t2)-px(t1)
      t2-t1
      px(2.00 s)
      =
      (1.40 kg·m/s2)(2.00 s) -(2.80 kg·m/s3)(2.00 s)2
      =
      -8.40 kg·m/s
      px(1.00 s)
      =
      (1.40 kg·m/s2)(1.00 s) -(2.80 kg·m/s3)(1.00 s)2
      =
      -1.40 kg·m/s
      Fx-av
      =
      (-8.40 kg·m/s)-(-1.40 kg·m/s)
      (2.00 s-1.00  s)
      =
      -7.00 N

      (d) [4 pts] Does the acceleration of the particle increase or decrease with time or just remain constant?
      The acceleration DECREASES with increasing time. It points in the negative x-direction.
      ax
      =
      Fx(t)
      m
      =
      a-2bt
      m
      ax(0 � t < a/2b)
      >
      0
      ax(t > a/2b)
      <
      0
FORMULAS
W
=
F·Dr
W
=
�
� 		F·dr
K
=
1
2
 mv2
Usp
=
1
2
 kx2
Wtot
=
Kf-Ki
E
=
K+U
F
=
ma
�
Fx
=
max
�
Fy
=
may
v2
=
v2o+2aDs
v
=
vo+at
J
=
Dp
=
m(vf-vi)
J
=
FavDt
Fnet
=
dpsys
dt
Favg
=
Dp
Dt


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On 02 Nov 2015, 17:15.