Download SPH4U Dynamics Test 5

SPH4U Physics Grade 12
Dynamics Test
Multiple Choice
Identify the letter of the choice that best completes the statement or
answers the question.
1. Which of the following is NOT considered to be one of the
“fundamental forces”?
a. gravity
d. weak nuclear
b. friction
e. electromagnetic
c. strong nuclear
2. The free-body diagram of a block being pushed up a rough ramp is
best represented by
a. A
b. B
c. C
d. D
e. E
3. An elevator is moving upward at a constant velocity. What is the
relationship between the gravitational force 𝐹⃗ g acting on the elevator and
the tension 𝐹⃗ T in the cable that supports the elevator?
a. |𝐹⃗ g| = |𝐹⃗ T|
d. |𝐹⃗ g| ≥ |𝐹⃗ T|
b. |𝐹⃗ g| > |𝐹⃗ T|
e. |𝐹⃗ g| ≤ |𝐹⃗ T|
c. |𝐹⃗ g| < |𝐹⃗ T|
4. Which of the following would be considered an “inertial” frame of
reference?
a. a moving escalator
b. a car moving through a turn at a constant speed
c. an object in “free fall”
d. a car pulling away as a traffic light turns green
e. all of the above
5. Imagine that you are travelling in a train and you have a drink sitting
on the dining table in front of you. The train suddenly stops and the
drink ends up in your lap. What force acting on the drink is responsible
for its sudden motion?
a. the force of the table acting on the drink
b. the force of the track on the wheels
c. the force of the wheels on the track
d. the force of the drink on itself
e. There is no force acting on the drink that is responsible for its motion.
Answers
1
B
2
E
3
A
4
A
5
E
Short Answer
1. Describe situations where the normal force and the gravitational force
acting on an object are equal. Describe situations where they are not
equal. (2 marks)
Answer
The normal force and gravitational force acting on an object are equal
when the object sits on a horizontal surface with nothing either pulling it
up or pushing it down. These two forces are not equal when the object is
not on a horizontal surface. In this case, the normal force is equivalent to
the component of gravity perpendicular to the plane. They would also
not be equal, even on a horizontal surface, if something was pushing the
object down or pulling it up somewhat. This could be the case for an
object with a rope tied to it, being pulled along a horizontal surface with
the rope at some angle to the plane.
Problems
1. A 12.0-kg box is pushed along a horizontal surface by a 24 N force as
illustrated in the diagram. The frictional force (kinetic) acting on the
object is 6.0 N.
(a) What is the acceleration of the object? (4 marks)
Solution
Free-body diagram: FN acting up
Fg acting down
FA acting as illustrated
FK acting to the right
“Up” and “to the right” are the positive directions.
Horizontally:
The acceleration of the object is 1.0 m/s2.
(b) Calculate the value of the normal force acting on the object. (3
marks)
Solution
Vertically:
The normal force is 1.3 ×102 N[up].
(c) If the 12.0-kg object then runs into a 4.0-kg object that increases the
overall friction by 3.0 N, what is the new acceleration? (3 marks)
Solution
Free-body diagram: FN acting up
Fg acting down
FA acting to the left
FK acting to the right
“Up” and “to the right” are the positive directions.
The acceleration of the two masses is 0.59 m/s2.
(d) What force does the 4.0-kg object exert on the 12.0-kg object when
the two are moving together? (3 marks)
Solution
Free-body diagram:
FN acting up
Fg acting down
FA acting to the left
FK acting to the right
F acting to the right (force of 4.0 kg object on 12.0 kg object)
“Up” and “to the right” are the positive directions.
The 4.0-kg object exerts a force of 5.3 N on the 12.0-kg object.
2. Two masses, 4.0 kg and 6.0 kg, are connected by a “massless” rope
over a “frictionless” pulley as pictured in the diagram. The ramp is
inclined at 30.0º and the coefficient of kinetic friction on the ramp is
0.18.
(a) Draw free-body diagrams of both masses. (2 marks)
Solution
For the 4.0-kg mass:
Free-body diagram:
FN acting perpendicular to the ramp (upward)
Fg acting down
FT acting up along the ramp (this is the positive direction)
FK acting down along the ramp (this is the negative direction)
For the 6.0-kg mass:
Free-body diagram:
Fg acting down (this is the positive direction)
FT acting up (this is the negative direction)
(b) Determine the acceleration of the system once it begins to slide. (4
marks)
Solution
For the 4.0-kg mass:
4.0 kg(a) = FT –𝜇 Kmg (cos 𝜃) – mg (sin 𝜃)
4.0 kg (a) = FT – 6.1 N – 19.6 N
4.0 kg(a) = FT – 25.7 N
For the 6.0-kg mass:
6.0 kg(a) = 58.8 N – FT
Solving the system of equations:
a = 3.3 m/s2
The acceleration of the 4.0-kg mass along the ramp is 3.3 m/s2.
(c) Determine the tension in the rope. (3 marks)
Solution
FT = 4.0 kg (a) +25.7 N
= 4.0 kg (3.31 m/s2) + 25.7 N
FT = 39 N
The tension in the cable is 39 N.
3. A pilot of mass 75 kg takes her plane into a dive, pulling out of it
along a circular arc as she nears the ground. If the plane is flying at 1.5
× 102 km/h along the arc, what is its radius such that the pilot feels four
times heavier than normal? Provide an appropriate free-body diagram. (4
marks)
Solution
The free body diagram of the pilot at the bottom of the arc:
FN = force of seat exerted upward on the pilot (the normal force)
FN = 4mg
The radius of the arc is 59 m.
4. A rock of mass 4.0 ×102 g is tied to one end of a string that is 2.0 m in
length. Holding the other end above his head, a boy swings the rock
around in a circle whose plane is parallel to the ground.
(a) If the string can withstand a maximum tension of 4.5 N before
breaking, what angle to the vertical does the string reach just before
breaking? (4 marks)
Solution
The centripetal acceleration (and force) are directed toward the centre of
the circle. This means that the two vertical forces are balanced.
Fg = FT sin 𝜃
In the horizontal plane, the centripetal force is supplied by the horizontal
component of the tension:
FC = FTcos 𝜃
FT = 4.5 N
Fg = mg = 0.400 kg(9.8 N/kg)
Fg = 3.92 N
𝜃 = 61º
The angle to the vertical is 29°.
(b) At what speed is the rock travelling just as the string breaks? (4
marks)
Solution
The speed of the rock is 2.3 m/s.
2. ANS:
(a)
(b)
(c)