INTRODUCTORY PHYSICS TEST NO. 1 SPRING 08 NAME KEY
1. A scalar has _______.
(a) neither magnitude nor direction (b) magnitude but not direction
(c) direction but not magnitude (d) both magnitude and direction
2. A vector has _______.
(a) neither magnitude nor direction (b) direction but not magnitude
(c) magnitude but not direction (d) both magnitude and direction
3. The SI units of acceleration are
(a) m (b) m/s/s (c) s (d) m/s
4. Tommy throws a baseball straight up in the air. At the highest point
in its path
(a) both its velocity and acceleration are zero.
(b) its velocity is zero but its acceleration is 9.8 m/s/s downwards.
(c) both its velocity and acceleration are nonzero.
(d) its acceleration is zero but its velocity is not.
5. A skier in the super-g leaves the starting gate and, 12 s later,
reaches a speed of 36 m/s. What was her average acceleration?
(a) 3.6 m/s/s (b) 4.8 m/s/s (c) 0.333 m/s/s (d) 3 m/s/s
6. Mass is the measure of a stationary object's _______.
(a) inertia (b) weight (c) size (d) density
7. If a constant net force is acting on an object, it will move with
(a) constant velocity. (b) increasing acceleration.
(c) decreasing acceleration. (d) constant acceleration.
8. An ice skater stands on smooth ice holding a basketball. He throws
the ball to a friend. Why doesn't Newton's third law require he
recoil backward at the same speed as the ball goes forward?
(a) The force on the ball is much greater than the reaction force.
(b) The mass of the skater is much greater than that of the ball.
(c) The reaction force only operates on the skater's hands.
(d) The skates were undoubtedly frozen to the ice.
9. Weight is
(a) the force of gravity acting on a body.
(b) the same thing as mass.
(c) the measure of an object's inertia.
(d) independent on where the object happens to be.
10. The condition for ROTATIONAL equiblibrium of a body is that
(a) no net force acts on it. (b) no forces act on it.
(c) no net torque acts on it. (d) no torques act on it.
11. Which of the following is NOT true of torque?
(a) It is the same thing as force.
(b) It is defined as force times lever arm.
(c) It is increased by increasing the lever arm.
(d) It is the cause of rotational motion.
12. An object with a mass of 10 kg weighs 98 N on the surface of the
earth. On the moon the pull of gravity is only 1/6 of that on the
earth. What is the mass and weight, respectively, of the same object
on the surface of the moon?
(a) 10 kg, 98 N (b) 1.7 kg, 98 N (c) 1.7 kg, 16 N (d) 10 kg, 16 N
13. Consider a seesaw that can be balanced upon its pivot with no one
riding it. If a 30 lb child gets on one side 4 ft from the pivot,
what weight child would be required for balance sitting 3 ft from the
pivot on the other side?
(a) 20 lb (b) 40 lb (c) 60 lb (d) 120 lb
14. What is the minimum net force you can apply to an object by applying
two equal forces of 100 lb?
(a) 0 lb (b) 50 lb (c) 100 lb (d) 200 lb
15. The work done by a force acting on an object is defined as
(a) the force times the time interval through which it acts.
(b) the force times the distance the object moves.
(c) the force times the speed with which the object moves.
(d) the force times its lever arm.
16. Which of the following is NOT a definition of power?
(a) the amount of work done over given distance
(b) the energy expended per unit time
(c) the amount of work done per unit time
(d) the amount of energy supplied per unit time
17. Energy
(a) is the same thing as force, the more force the more energy.
(b) is the same thing as work.
(c) is imparted to an object by doing positive net work on the object.
(d) only comes in one form, the energy of motion.
18. A box is pushed across the floor. After being released it slows down
and comes to rest. After the box was released
(a) mechanical energy (KE + PE) was conserved.
(b) kinetic energy only was conserved.
(c) total energy (KE + PE + heat) was conserved.
(d) the total energy decreased, disappearing as the box slowed down.
19. The work done by friction on an object sliding over a surface is ___.
(a) zero (b) positive (c) negative
20. Salesman Joe says his machine can output twice as much energy as is
put into it, whereas salesman Sam claims his machine can output twice
as much force as is put into it. Who do you KNOW is lying?
(a) both of them (b) neither of them (c) Sam (d) Joe
21. An automobile is traveling at a speed of 80 km/h, when a strong
headwind arises. The driver does not compensate by pressing down
harder on the accelerator pedal and the automobile slows to 40 km/h,
since the drive train is supplying the same amount of force as before.
The mechanical power expended by the car is _______.
(a) the same as before (b) half as great as before
(c) twice as great as before (d) four times as great as before
22. In its broadest meaning the conservation of energy states that
(a) Heat energy can neither be created nor destroyed.
(b) All energy ends up as heat.
(c) Energy can be transformed from one type to another but cannot be
created or destroyed.
(d) Mechanical energy cannot be created or destroyed.
23. For mechanical energy to be conserved
(a) kinetic energy but not potential energy must be constant.
(b) potential energy but not kinetic energy must be constant.
(c) the force of gravity must be absent.
(d) friction must be absent.
24. A contest of strength involves pushing a heavy sled a certain distance
and across a finish line. Bubba, by pushing with twice the force
as his nearest competitor, Butch, gets the sled across the finish
line in half Butch's time. How much greater was Bubba's power
output than Butch's? (Note: They pushed the cart the same distance.)
(a) twice as much (b) three times as much
(c) four times as much (d) eight times as much
25. Seat belts are often designed to give a little before they become
taut. This is to
(a) allow passengers room to move around while riding in the car.
(b) enable passengers to be thrown clear of the collision in some
types of accidents.
(c) allow passengers to escape from a burning car.
(d) reduce the acceleration of the passenger, and thus the force
acting on the passenger, during a collision.
Problems
1. In shuffleboard the object is to push a disk to give it just enough
speed to get it to go a certain distance. You have observed that
when a player gives a disk just the right amount of speed to go 8.0 m,
it takes 4.0 s for the disk to travel the 8.0 m distance before coming
to rest. For the following assume constant acceleration. (a) What
is the average velocity of the disk as it travels the 8.0 m distance?
(b) What initial velocity must a player give the disk to travel
exactly this distance? (c) What is the (negative) acceleration of
the disk as it travels?
(a) vavg = s/t = 8.0 m / 4.0 s = 2.0 m/s
(b) For constant acceleration, the average velocity is also given
by
vavg = (vo + v)/2.
In this case the final velocity, v, is zero, and you already
know that the average velocity is 2.0 m/s. Solving for the
initial velocity, vo,
vo = 2vavg = 2(2.0 m/s) = 4.0 m/s
(c) There are several formulas you can now use to get acceleration.
The most straightforward way is probably to use
a = (v - vo)/t = (0 - 4.0 m/s)/(4.0 s) = -1.0 m/s2.
2. In a hypothetical example, suppose an astronaut were to throw a rock
on the moon straight up. Further, suppose that the initial velocity
of the rock were 20 m/s. On the surface of the moon the acceleration
of gravity is 1.6 m/s/s. Find the following. (a) The velocity of
the rock after 2.0 s. (b) The maximum height reached by the rock.
The equation for velocity versus time given a constant
acceleration is v = vo + at, where "a" is the
acceleration, t is the elapsed time for the motion, and
vo is the initial velocity. If you take up
to be positive, then vo = 30 m/s and a = -1.6 m/s2.
This makes the equation read
v = 20 m/s + (-1.6 m/s2)t.
For t = 2.0 s,
v = 20 m/s + (-1.6 m/s2)(2 s) = 16.8 m/s.
Maximum height can be gotten three ways. One is to solve for
the time when the rock reaches its highest point. At this
point its velocity will be zero, since it is at the "turning
around" point. The equation for the velocity reads, for this
case,
0 = 20 m/s + (-1.6 m/s2)t,
where t is the unknown. Solving for t you get
t = -(20 m/s)/(-1.6 m/s2) = 12.5 s.
Now you can use the equation of position versus time,
s = vot + (1/2)at2,
to compute the height s, given that vo = 20 m/s,
a = -1.6 m/s2, and t = 12.5 s. The result is
s = (20 m/s)(12.5 s) + (1/2)(-1.6 m/s2)(12.5 s)2 = 125 m.
Or, you can use the equation
s = vavgt
where vavg is found from vavg = (vo + v) / 2,
and the "final" velocity v = 0 at the highest point.
s = [(20 m/s + 0)/2](12.5 s) = 125 m.
A third way to find the height reached is to use the equation
v2 = vo2 + 2as.
The advantage of using this equation is that you don't need to
solve for the time first. Putting the known values into this
equation you get
0 = (20 m/s)2 + 2(-1.6 m/s2)s.
Solving for s,
s = -(20 m/s)2/(-3.2 m/s2) = 125 m.
3. A dynamics cart in a physics lab is propelled down a track by means
of a fan attachment. The dynamics cart plus fan weighs 550 g. The
fan exerts a force of 0.10 N on the cart, and there is also a force
of friction between the cart and the track of 0.020 N. When the fan
is turned on the cart moves along the track. (a) What is the net
force acting on the cart? (b) What is the acceleration of the cart?
(c) The track is about 80 cm long. How much time does it take the
cart to move 80 cm if it starts from rest?
Taking the force of the fan as positive, the force of friction
will be negative. The net force is thus
Fnet = 0.10 N - 0.02 N = 0.08 N.
Newton's second law states that the acceleration equals the
net force divided by the mass that is being accelerated,
or
a = Fnet/m = 0.08 N / 0.55 kg = 0.145 m/s2
The equation of the position of a constantly accelerating object
versus time is
s = vot + (1/2)at2.
For this problem s = 0.80 m, vo = 0 (starting from rest), and
a = 0.145 m/s2. Therefore
0.80 m = (1/2)(0.145 m/s2)t2.
Solving for t,
t = [2(0.80 m)/0.145 m/s2]1/2 = 3.3 s.
(Recall that taking the square root and raising to the 1/2 power
are the same thing.)
4. Jesse built a "Pinewood Derby" car with a mass of 150 g to race
against other such cars at a Boy Scout meeting. The cars are
released at the top of a track that curves down and finally runs
horizontally into a box filled with soft cloth to brake the cars to
a stop. The top of the track is about 1.8 m above the point where
the cars' motion becomes horizontal. (a) What was the gravitational
potential energy of Jesse's car at the starting point, taking the
horizontal section of the track as the zero level? (b) Assuming
friction can be neglected, use the conservation of energy to find the
speed of the car when it reached the horizontal section of the track.
Since the car is 1.8 m above the zero level and has a mass
of 0.15 kg, the potential energy is
PE = mgh = (0.15 kg)(9.8 m/s2)(1.8 m) = 2.65 J.
If no friction acts, the PE of the car will be converted to KE
during its descent, meaning the car must have a KE = 2.65 J at
the zero level. KE = (1/2)mv2, so, solving for v,
v = (2KE/m)1/2 = [2(2.65 J)/(0.15 kg)]1/2 = 5.9 m/s.
5. A sports car with a mass of 1500 kg accelerates from 0 to 32 m/s in
8.0 s. (a) What are the acceleration of the car and the net force
acting on it? (b) How far did the car travel in the 8.0 s? (c) What
was the work done by the net force on the car? (d) Calculate the
final kinetic energy of the car and show that it is equal to the work
done by the net force.
The acceleration can be found from the initial and final velocities
and the time elapsed.
a = (v - vo)/t = (32 m/s - 0)/(8.0 s) = 4.0 m/s2.
The net force is Fnet = ma,
Fnet = (1500 kg)(4.0 m/s2) = 6000 N.
Using an equation seen earlier in this key,
s = [(vo + v)/2]t = [(0 + 32 m/s)/2](8.0 s) = 128 m.
The work done on the car by the net force can most easily be
computed from
Wnet = Fnets = (6000 N)(128 m) = 7.68 × 105 J.
The KE is
KE = (1/2)mv2 = (1/2)(1500 kg)(32 m/s)2 = 7.68 × 105 J,
which is equal to the net work done as required by the work-energy
theorem.