1201W.200 Sample quiz 1 Fall 2010

1201W.200 Sample quiz 1 Fall 2010
University of Minnesota
M.Zudov
This test carries 25 points which will be combined with individual test to give a grand
total of 100 points. The acceleration due to gravity is 9.8 m/s2 = 32 ft/s2; 1 mile = 5280 ft
1.g. A child puts a brick of mass m2 on top of a brick of mass m1 and pushes them up an
inclined plane by exerting a constant force F parallel to the plane on the lower brick, as
shown in the diagram. The plane is at an angle θ to the horizontal and there is no
frictional force between the plane and the lower brick m1. The coefficient of static friction
between the two bricks is μs > tan θ. If the two bricks are to move together without
slipping find:
(a) corresponding acceleration of the blocks a
(b) the normal force N1 exerted by the plane on brick m1
(c) the frictional force between the blocks
(d) the maximum force F that the child can exert
Bonus part: How would your results change if the force F is applied in the opposite
direction, i.e. if the child pushes the bricks down the plane?
m2
m1
θ , m1 , m2 , μ s , a1 = a 2 = a; f = f max = μ s N 2
No friction
F
a − ? N 1 − ? f − ? Fmax − ?
θ
Pushing up :
⎧m a = − m2 g sin θ + f
(2) ⇒ ⎨ 2
⎩ 0 = − m2 g cos θ + N 2
⎧m a = − m1 g sin θ + F − f
(1) ⇒ ⎨ 1
⎩ 0 = − m1 g cos θ − N 2 + N 1
(2) x + (1) x ⇒ (m1 + m2 )a = F − (m1 + m2 ) g sin θ ⇒ a =
F
− g sin θ
m1 + m2
(2) y + (1) y ⇒ 0 = −(m1 + m2 ) g cos θ + N 1 ⇒ N 1 = (m1 + m2 ) g cos θ
(2) x ⇒ f = m2 (a + g sin θ ) ⇒ f =
F = Fmax =
m2
F
m1 + m2
m1 + m2
m + m2
f max = 1
μ s m2 g cos θ ⇒ Fmax = (m1 + m2 ) μ s g cos θ
m2
m2
1201W.200 Sample quiz 1 Fall 2010
University of Minnesota
M.Zudov
Problems 1 and 2 carry 25 points each and the five multiple choice problems carry 5
points each. Attempt all problems. Work problems 1 and 2 in the space provided; the
proctor will give you extra paper if you need it. Give the answers to the multiple choice
problems on the General Purpose Answer Sheet or “Bubble Sheet” provided. Put your
name, ID number and the name of your TA on each page. The acceleration due to gravity
is 9.8 m/s2 = 32 ft/s2; 1 mile = 5280 ft
1201W.200 Sample quiz 1 Fall 2010
Name:
ID No:
(last name
University of Minnesota
M.Zudov
TA Name:
first name)
1.1 A place kicker wants to kick a field goal. The uprights are x = 40.0 m away and the
horizontal bar is y = 4.00 m above the ground. The ball was kicked at an angle of θ =
45.0˚ to the horizontal and it hit the horizontal bar
(a) What was the speed with which the ball was kicked?
(b) How long did it take the ball to reach the uprights?
(c) With what speed did the ball hit the uprights?
x0 = y 0 = 0; v0 x = v0 ⋅ cos θ ; v 0 y = v0 ⋅ sin θ ;θ = 40°; x = 40 m; y = 4 m; a x = 0; a y = − g ;
(a) v0 − ?
(b) t − ?
(a) v − ?
x = 0 + v0 x t ⇒ t = x v0 x ⇒
y = 0 + v0 y t −
v0 y 1 2
1 2
1
− gt = x tan θ − gt 2 ⇒ t =
gt = x
2
2
v0 x 2
x = v0 x t ⇒ v0 x = x t ⇒ v0 =
2( x tan θ − y )
=
g
2 ⋅ (40 ⋅ 1 − 4)
= 2.71 s
9.8
40
x
=
= 20.9 m/s
t cos θ 2.71 ⋅ 0.707
v y2 = v 02y − 2 gy ⇒ v 2 = v y2 + v x2 = v02 sin 2 θ − 2 gy + v 02 cos 2 θ = v02 − 2 gy ⇒ v = 436.8 − 78.4 = 18.9 m/s
1201W.200 Sample quiz 1 Fall 2010
University of Minnesota
M.Zudov
Name:
ID No:
(last name
TA Name:
first name)
1.2. As shown in the sketch, two blocks are connected by a light string which runs over a
frictionless, massless pulley. Block 1, of mass 20 kg, slides on an inclined plane whose
angle is 20˚ and the coefficient of kinetic friction is 0.15. Block 2 moves vertically.
(a) If block 1 slides down the inclined plane at constant speed, what is the mass of block 2?
(b) If block 1 slides up the inclined plane at constant speed, what is the mass of block 2?
1
2
20˚
θ = 20°, m1 = 20 kg, μ k = 0.15, a1 = a 2 = 0
m2 − ?
⎧0 = m1g + N + f + T1 (1)
ma = 0 = ∑ Fi ⇒ ⎨
(2)
i
⎩0 = m2 g + T2
(a) down :
⎧0 = m1 g sin θ − μ k N − T
(1) ⇒ ⎨
⇒ T = m1 g (sin θ − μ k cos θ )
⎩0 = − m1 g cos θ + N
(2) ⇒ 0 = m2 g − T ⇒ m2 = T g = m1 (sin θ − μ k cos θ ) = ... = 4.02 kg
(b) up (just need to change the sign of the frictional force) :
⎧0 = m1 g sin θ + μ k N − T
(1) ⇒ ⎨
⇒ T = m1 g (sin θ + μ k cos θ )
⎩0 = − m1 g cos θ + N
(2) ⇒ 0 = m2 g − T ⇒ m2 = T g = m1 (sin θ + μ k cos θ ) = ... = 9.66 kg
1201W.200 Sample quiz 1 Fall 2010
University of Minnesota
M.Zudov
1.m Multiple choice answers are marked on the “Bubble Sheet”
(1) In terms of unit vectors i pointing along the x-axis and j pointing along the y-axis, a
G
vector A = i- 3 j. If it is specified instead by the magnitude (A=2) and the angle θ it
makes with the positive x-axis, the value of θ is:
− 3
tan θ =
⇒ θ = −60° = 300°
1
(a) 30˚ (b) 60˚ (c) 120˚ (d) 210˚ (e) 300˚
(2) A car moving with an initial velocity of 100 m/s north has a constant acceleration of
10 m/s2 south. After 6 seconds its velocity will be:
(a) 60 m/s north
(b) 60 m/s south
(c) 160 m/s north
(d) 40 m/s north
(e) 40 m/s south
(3) A 1 kg stone is attached to a string of length 1.25 m and swung in a vertical circle (the
plane of the circle is vertical). At the top of the circle the string has a speed of 3.5 m/s and
the tension in the string is close to:
(a) zero (b) 9.8 N (c) 4.9 N (d) 12.25 N (e) the stone cannot be moving in a circle
ma = m
v2
= mg + T ⇒ T = m(v 2 r − g )
r
(4) As shown in the picture, a pendulum bob which weight 1 N is held at rest at an angle
θ from the vertical by a 2 N horizontal force F. The tension in the string supporting the
pendulum bob is:
(a) cos θ N
(b) 2 / cos θ N
(c) 5 N
θ
(d) 1 N
(e) 2 N
F
(5) For an object moving at a constant velocity, which of the following statement is
always true?
(a) A constant force is needed to keep the object moving at a constant velocity.
(b) For the object to move at a constant velocity, no forces can be acting on it.
(c) The object is moving in circular motion.
(d) None of the above