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The first midterm quiz is Monday, February 16
at 1:25pm -2:15pm in the following rooms
133 Tate if your last name starts with A-L
170 Tate if your last name starts with M-Z
(basement level)
Closed book, notes. No computers, calculators,
or cell phones.
Review
velocity is (change in position)/time elapsed
acceleration is (change in velocity/time elapsed
Changes can be in magnitude or in direction or both
total force =mass x acceleration
Kinds of forces: gravitational,electromagnetic
Different kinds of forces can add to give total force.
Near the earth’s surface
gravitational force = Mass x g down
(1/2)mv2 + mgh = constant (conservation of energy)
kinetic
gravitational
potential
Review page 2:
When one object exerts a force F on another object
through a distance d along the direction of the
force, an amount of WORK Fd is done by the
first object on the second and an amount of
energy Fd is transferred from the first object to
the second.
Newton’s third law says that when one object
exerts a force F on a second object, then the second
object exerts a force –F(same magnitude, opposite
direction) on the first object.
Some practise questions.
Suppose the 1 kg block being pulled across the table
is accelerating at 8m/s2 and that the force pulling the
block is 10 newtons. What is the magnitude and
direction of the friction force on the block?
A. 9.8 newtons down
B. 10 newtons backward
C. 2 newtons backward
D. 2 newtons forward
Answer C.
ma=1kg x 8m/s2 = 10 newtons -magnitude of backward friction force
If the same 1kg block (pulled with a 10 newton force)
was pulled through 1/2 meter while accelerating at 8
m/s2 how much thermal energy was generated?
A. 5 joules
B. 4 joules
C. 1 joule
D. 4.9 joules
Answer C.
Friction force on the table x ½ meter =1joule
Notes:
Total work done by pulling the block =
10 newtons x ½ meter = 5 joules
So the kinetic energy of the block
after 1 s must have been 4 joules.
See if you can show that the final kinetic
energy of the block is 4 joules
by finding the final velocity of the
block and using the definition
KE = ½ mv2
If you assume that the moon moves in a circle around the earth at
constant speed (which is approximately true), what is the direction of
the velocity, acceleration and total force on the moon at any
moment?
v
A.tangent to the circle
earth
a
straight toward earth
earth
F
straight toward
B. tangent
tangent
toward earth
C. toward earth
toward earth
toward earth
D. tangent
toward earth
tangent
Answer: A.
When I drop an elastic ball and it hits the floor
and bounces, what is the
direction of its velocity, acceleration and the total
force on it when
it is at its lowest point (in contact with the floor)?
v
a
F
A.
d
d
d
B.
0
u
d
C.
0
u
u
D.
u
u
u
E.
0
d
d
Answer: C.
When I dropped a rubber ball from one meter, it
was observed to rise to
0.75m after the first bounce. How was energy
conserved?
Answer:
A. Kinetic energy of ball increased.
B. The thermal energy of the floor and the ball
increased.
C. The kinetic energy of the earth increased.
D. The ball was spinning.
Answer B.
I push a cart up with a constant force up a slope to a height of 1/4
meter. The mass of the cart is 1/4 kg. The cart moves 1/2 meter
along the table as I do this. After I let it go, the cart continues up the
slope
until it reaches a height of 1/2 meter and then starts back down.
What was the magnitude of the force with which I pushed the cart?
A. 2.45/√5 newtons
B. 4.9/ √5 newtons
C. 4.9 newtons
D. 2.45 newtons
B.
Total energy gain =(1/4kg)x(9.8m/s2)x(1/2 m)
=Fx (distance along force)
=Fx√((1/2)2 + (1/4)2)=Fx(√5/4)
Solve for F
In the same situation, at what speed was the car
moving when it left my hand?
A. √4.9
B. √2.45
m/s
m/s
C. √(4.9√5)
D. √9.8
m/s
m/s
Answer A.
(¼ m)x(9.8m/s2)(1/4 kg)=(1/2)(1/4 kg) v2
Solve for v.
A baseball player hits a ball. The velocity as the ball leaves the bat
is 20 m/s up and 40 m/s horizontal. How much time does it take for
the ball to reach its highest point?
Answer:
A. 40/9.8 sec
B. 20√5/9.8 sec
C. 20/9.8 sec
D. 9.8/20 sec
Answer: C.
Ball loses 20m/s in the time t while decelerating
at 9.8m/s2 so
20m/s=(9.8m/s2 )xt
Solve for t.
Same baseball problem: How far away from the
batter in the horizontal
direction does the ball land?
A. 800/9.8m
B. 1600/9.8 m
C. 400/9.8 m
D. 3200/9.8 m
Answer: B
Time x horizontal velocity
Time is twice time to get to the highest point.