section a - multiple choice
for each question, choose the correct/best answer and place your solutions in the table below. only
the table will be graded, not any of the circled responses.
1. to determine average velocity, one would divide __________ by time.
a. distance
b. velocity
c. acceleration
d. displacement
e. none of the above are correct
2. the slope of a position time graph would yield:
a. displacement
b. distance
c. velocity
d. acceleration
e. change in acceleration
3. a cyclist rides a bicycle 4.0 km west, then 3.0 km north. what is the cyclists distance travelled?
a. 7.0 km 37°n of w
b. 7.0 km 37° w of n
c. 5.0 km 37°n of w
d. 5.0 km 37° w of n
e. 7.0 km
4. a ball is thrown vertically downward with a speed of 4.4 m/s from the top of a large building. the stone strikes the surface of the earth 25s later. from what height above the surface was the ball thrown?
a. 3.0×10³ m
b. 3.2 × 10³ m
c. 3000 m
d. 2400 m
e. 2800 m
5. which of the following below has an x component of 2.7 m w and a y component of 3.5 m n
a. 4.4 m n 52°w
b. 4.4 m w 52° n
c. 6.2 m nw
d. 0.80 m w 28°s
e. none of the above
6. in which of the following scenarios could kinetic friction be beneficial for safety?
a. pulling a heavy box across the floor
b. pulling up in a car to a stop sign
c. tobogganing down a hill
d. turning a door knob
e. none of the above
7. identical books a and b are pushed with horizontal forces along a rough table top. both books travel at constant velocities, but a’s speed with double b’s speed. which statement must be true?
a. the force of friction acting on a is double the force of friction acting on b
b. the applied force on a is double the applied force on b
c. the coefficient of friction between the table and book a is double the coefficient of friction between the table and book b
d. the net force on a is double the net force on b
e. none of the above
8. which of the following scenarios must be true about a non - zero net force?
a. the applied force must be zero
b. the force of friction must be zero
c. the object is moving at a constant velocity
d. the acceleration of the object is not zero
e. the direction of the applied force and the force of friction is the same
9. a 37.0 kg child sits on a light sled at rest. if the coefficient of static friction between the sled and the snow is 0.510, how much force is needed to start moving the sled and child?
a. 18.9 n
b. 185 n
c. 37.5 n
d. 180 n
e. none of the above
10. select the choice to make a true statement: the force of gravity between two objects is...
a. inversely proportional to r²
b. proportional to r²
c. inversely proportional to r
d. proportional to r
e. none of the above

Question

section a - multiple choice
for each question, choose the correct/best answer and place your solutions in the table below. only
the table will be graded, not any of the circled responses.
1. to determine average velocity, one would divide __________ by time.
   a. distance
   b. velocity
   c. acceleration
   d. displacement
   e. none of the above are correct
2. the slope of a position time graph would yield:
   a. displacement
   b. distance
   c. velocity
   d. acceleration
   e. change in acceleration
3. a cyclist rides a bicycle 4.0 km west, then 3.0 km north. what is the cyclists distance travelled?
   a. 7.0 km 37°n of w
   b. 7.0 km 37° w of n
   c. 5.0 km 37°n of w
   d. 5.0 km 37° w of n
   e. 7.0 km
4. a ball is thrown vertically downward with a speed of 4.4 m/s from the top of a large building. the stone strikes the surface of the earth 25s later. from what height above the surface was the ball thrown?
   a. 3.0×10³ m
   b. 3.2 × 10³ m
   c. 3000 m
   d. 2400 m
   e. 2800 m
5. which of the following below has an x component of 2.7 m w and a y component of 3.5 m n
   a. 4.4 m n 52°w
   b. 4.4 m w 52° n
   c. 6.2 m nw
   d. 0.80 m w 28°s
   e. none of the above
6. in which of the following scenarios could kinetic friction be beneficial for safety?
   a. pulling a heavy box across the floor
   b. pulling up in a car to a stop sign
   c. tobogganing down a hill
   d. turning a door knob
   e. none of the above
7. identical books a and b are pushed with horizontal forces along a rough table top. both books travel at constant velocities, but a’s speed with double b’s speed. which statement must be true?
   a. the force of friction acting on a is double the force of friction acting on b
   b. the applied force on a is double the applied force on b
   c. the coefficient of friction between the table and book a is double the coefficient of friction between the table and book b
   d. the net force on a is double the net force on b
   e. none of the above
8. which of the following scenarios must be true about a non - zero net force?
   a. the applied force must be zero
   b. the force of friction must be zero
   c. the object is moving at a constant velocity
   d. the acceleration of the object is not zero
   e. the direction of the applied force and the force of friction is the same
9. a 37.0 kg child sits on a light sled at rest. if the coefficient of static friction between the sled and the snow is 0.510, how much force is needed to start moving the sled and child?
   a. 18.9 n
   b. 185 n
   c. 37.5 n
   d. 180 n
   e. none of the above
10. select the choice to make a true statement: the force of gravity between two objects is...
    a. inversely proportional to r²
    b. proportional to r²
    c. inversely proportional to r
    d. proportional to r
    e. none of the above

Accepted Answer

### Question 1
# Brief Explanations:
Average velocity is defined as displacement divided by time. Distance is used for average speed, velocity is not divided by time here, and acceleration is unrelated. So the correct option is d.
# Answer:
d. displacement

### Question 2
# Brief Explanations:
The slope of a position - time graph is calculated as the change in position (displacement) divided by the change in time. By the definition of velocity ($v=\frac{\Delta x}{\Delta t}$), this slope gives velocity. Displacement is the change in position, distance is a scalar related to total path, acceleration is the slope of a velocity - time graph.
# Answer:
c. velocity

### Question 3
# Brief Explanations:
Distance traveled is the sum of the magnitudes of each segment of the path. The cyclist rides 4.0 km west and then 3.0 km north. So distance $d = 4.0\ km+3.0\ km = 7.0\ km$. The options with direction are for displacement, not distance.
# Answer:
e. 7.0 km

### Question 4
# Explanation:
## Step 1: Identify the kinematic equation
We use the kinematic equation for vertical displacement with initial velocity $u$, time $t$, and acceleration $g$ (acceleration due to gravity, $g = 9.8\ m/s^{2}$). The equation is $h=ut+\frac{1}{2}gt^{2}$.
## Step 2: Substitute the values
Given $u = 4.4\ m/s$, $t = 25\ s$, $g=9.8\ m/s^{2}$.
First, calculate $ut=(4.4\ m/s)	imes25\ s = 110\ m$.
Then, calculate $\frac{1}{2}gt^{2}=\frac{1}{2}	imes9.8\ m/s^{2}	imes(25\ s)^{2}=\frac{1}{2}	imes9.8	imes625 = 3062.5\ m$.
## Step 3: Calculate the total height
$h=110\ m + 3062.5\ m=3172.5\ m\approx3.2	imes 10^{3}\ m$.
# Answer:
b. $3.2	imes 10^{3}\ m$

### Question 5
# Explanation:
## Step 1: Calculate the magnitude of the vector
The magnitude of a vector with $x$ - component $x = 2.7\ m$ (west) and $y$ - component $y = 3.5\ m$ (north) is given by $r=\sqrt{x^{2}+y^{2}}$.
Substitute $x = 2.7\ m$ and $y = 3.5\ m$: $r=\sqrt{(2.7)^{2}+(3.5)^{2}}=\sqrt{7.29 + 12.25}=\sqrt{19.54}\approx4.4\ m$.
## Step 2: Calculate the angle
The angle $	heta$ (measured from the north axis towards the west) is given by $	an	heta=\frac{x}{y}$.
$	an	heta=\frac{2.7}{3.5}\approx0.771$, so $	heta=\arctan(0.771)\approx37^{\circ}$? Wait, no, let's recalculate. Wait, if $x = 2.7$ (W) and $y = 3.5$ (N), the angle with respect to the north axis is $	heta=\arctan(\frac{x}{y})=\arctan(\frac{2.7}{3.5})\approx37^{\circ}$? Wait, no, $	an	heta=\frac{opposite}{adjacent}=\frac{2.7}{3.5}$, so $	heta=\arctan(\frac{2.7}{3.5})\approx37^{\circ}$, but the direction is $N\ 37^{\circ}W$? Wait, no, let's check the options. Option a is $4.4\ m\ [N\ 52^{\circ}W]$? Wait, maybe I made a mistake. Wait, $	an	heta=\frac{x}{y}=\frac{2.7}{3.5}\approx0.77$, so $	heta\approx37^{\circ}$, but the angle with respect to the west axis: $	an\alpha=\frac{y}{x}=\frac{3.5}{2.7}\approx1.296$, $\alpha\approx52^{\circ}$. So the direction is $N\ 52^{\circ}W$ (because from north, we turn $52^{\circ}$ towards west). And the magnitude is $\approx4.4\ m$. So option a is correct.
# Answer:
a. $4.4\ m\ [N\ 52^{\circ}W]$

### Question 6
# Brief Explanations:
- Option a: Pulling a heavy box across the floor, kinetic friction opposes the motion and makes it harder, not beneficial for safety.
- Option b: Pulling up in a car to a stop sign - we use static friction when stopping a car (rolling friction and static friction for braking), not kinetic friction.
- Option c: Tobogganing down a hill, kinetic friction would slow down the toboggan, which can be beneficial for safety to prevent it from going too fast.
- Option d: Turning a door knob uses static friction, not kinetic friction.

So the correct option is c.
# Answer:
c. tobogganing down a hill

### Question 7
# Brief Explanations:
Both books move at constant velocity, so the net force on each is zero (Newton's first law). The force of friction $f=\mu N$, where $N$ is the normal force. Since the books are identical, $N$ (equal to weight) is the same, and the coefficient of friction $\mu$ is the same (same table and books). So the force of friction on both is the same. The applied force must equal the force of friction, so the applied force on both is the same. The coefficient of friction depends on the surfaces, not the speed. The net force on both is zero. So none of the statements a - d are true.
# Answer:
e. none of the above

### Question 8
# Brief Explanations:
From Newton's second law $F = ma$. If the net force $F
eq0$, then acceleration $a=\frac{F}{m}
eq0$.
- Option a: If net force is non - zero, applied force can't be zero.
- Option b: Friction can be non - zero and net force still non - zero (e.g., applied force greater than friction).
- Option c: If net force is non - zero, velocity is changing (acceleration is non - zero), so object can't move at constant velocity.
- Option e: Applied force and friction (when present) act in opposite directions for sliding objects.

So the correct option is d.
# Answer:
d. the acceleration of the object is not zero

### Question 9
# Explanation:
## Step 1: Calculate the normal force
The normal force $N$ on the sled and child is equal to their weight. The mass $m = 37.0\ kg$, so $N=mg$, where $g = 9.8\ m/s^{2}$. $N=37.0\ kg	imes9.8\ m/s^{2}=362.6\ N$.
## Step 2: Calculate the static friction force
The force needed to start moving is equal to the maximum static friction force $f_{s}=\mu_{s}N$. Given $\mu_{s}=0.510$, $f_{s}=0.510	imes362.6\ N\approx185\ N$.
# Answer:
b. 185 N

### Question 10
# Brief Explanations:
According to Newton's law of gravitation, the force of gravity between two objects is $F = G\frac{m_1m_2}{r^{2}}$, where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between their centers. So the force of gravity is inversely proportional to $r^{2}$.
# Answer:
a. Inversely proportional to $r^{2}$

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