Question 1

A ball is rolled horizontally off a table with an initial speed of 0.18 m/s.A stopwatch measures the ball's trajectory time from table to the floor to be 0.30 s.What is the height of the table? Note: g = 9.8 m/s² and air resistance is negligible. A) 0.11 m B) 0.22 m C) 0.33 m D) 0.44 m

Accepted Answer

D

Question 2

A 4000-N weight is suspended in equilibrium by two cables.Cable 1 applies a horizontal force to the right of the object and has a tension T₁.Cable 2 applies a force upward and to the left at an angle of 37.0° to the negative x axis and has a tension T₂.Find T_2. A) 4000 N B) 6650 N C) 8310 N D) 3340 N

Accepted Answer

First, let's draw a diagram of the system:

```
            ^ T2
            |
            |
            |
 T1 <-------O------->
            | 4000 N
            |
            |
           θ=37°
```

Let's start by resolving the forces in the horizontal direction:

```
T1 = 4000 N
T2 cos(37°) = T1
T2 = T1 / cos(37°) = 6221.32 N
```

Now, we can use the vertical equilibrium equation to find S1U1B0:

```
T2 sin(37°) + S1U1B0 = 4000 N
S1U1B0 = 4000 N - T2 sin(37°)
S1U1B0 = 4000 N - 3751.83 N
S1U1B0 = 248.17 N
```

Therefore, the answer is B) 6650 N.

Question 3

A baseball thrown from the outfield is released from shoulder height at an initial velocity of 39.2 m/s at an initial angle of 30.0° with respect to the horizontal.What is the maximum vertical displacement that the ball reaches during its trajectory?
A) 11.0 m
B) 19.6 m
C) 22.1 m
D) 44.1 m

Accepted Answer

Using the equation for maximum height of a projectile with initial velocity and launch angle, we have:
h_max = (v_i^2*sin^2(theta))/(2*g)
where v_i is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity. Plugging in the values given:
h_max = (39.2 m/s)^2 * sin^2(30.0°)/(2*9.81 m/s^2) = 19.6 m
So the maximum vertical displacement that the ball reaches during its trajectory is 19.6 meters.

Question 4

A trapeze artist,with swing,weighs 700 N;he is momentarily held to one side by his partner so that the swing ropes make an angle of 30.0° with the vertical.In such a condition of static equilibrium,what is the horizontal force being applied by the partner?
A) 924 N
B) 404 N
C) 808 N
D) 233 N

Accepted Answer

The horizontal force can be found using trigonometry. The tension in the rope has a vertical component equal to the weight (700 N) and a horizontal component equal to the force applied by the partner. Using the angle, the horizontal force is 700 N * tan(30°) = 404 N.

Question 5

A 200-N crate rests on a ramp;the maximum angle just before it slips is 21° with the horizontal.What is the coefficient of static friction between crate and ramp surfaces?
A) 0.11
B) 0.21
C) 0.38
D) 0.47

Accepted Answer

The coefficient of static friction (μ) is equal to the tangent of the angle of repose (θ), so μ = tan(21°) ≈ 0.38.

Question 6

A baseball thrown from the outfield is released from shoulder height at an initial velocity of 39.2 m/s at an initial angle of 30.0° with respect to the horizontal.If it is in its trajectory for a total of 4.00 s before being caught by the third baseman at an equal shoulder-height level,what is the ball's net vertical displacement during its 4.00-s trajectory?
A) 11.0 m
B) 9.80 m
C) 22.1 m
D) zero

Accepted Answer

The net vertical displacement is zero because the ball is caught at the same height from which it was thrown.

Question 7

A 500-N tightrope walker stands at the center of the rope.If the rope can withstand a tension of 1400 N without breaking,what is the minimum angle the rope can make with the horizontal?
A) 4°
B) 8°
C) 10°
D) 15°

Accepted Answer

The tension in the rope must be equal to the weight of the tightrope walker in order to maintain equilibrium. Therefore:

Tension = Weight = 500 N

The maximum tension the rope can withstand without breaking is 1400 N. We can use this information to find the minimum angle the rope can make with the horizontal as follows:

Tension = Weight / sinθ

1400 N = 500 N / sinθ

sinθ = 500 N / 1400 N

θ = sin^-1(500 N / 1400 N)

θ ≈ 10°

Therefore, the minimum angle the rope can make with the horizontal is approximately 10°, which corresponds to choice (C).

Question 8

A ball is rolled horizontally off a table with an initial speed of 0.18 m/s.A stopwatch measures the ball's trajectory time from table to the floor to be 0.30 s.How far away from the table does the ball land? Note: g = 9.8 m/s² and air resistance is negligible. A) 0.054 m B) 0.072 m C) 1.2 m D) 1.9 m

Accepted Answer

The correct answer is A because the ball's horizontal velocity is constant, and the vertical velocity is due to gravity. The horizontal distance traveled is equal to the initial horizontal velocity multiplied by the time of flight. The time of flight is equal to the vertical distance traveled divided by the vertical velocity. The vertical distance traveled is equal to half the acceleration due to gravity multiplied by the time of flight squared.

Question 9

A trapeze artist,with swing,weighs 700 N;he is being held to one side by his partner so that the swing ropes make an angle of 30.0° with the vertical.In such a condition of static equilibrium what is the tension in the rope?
A) 924 N
B) 404 N
C) 808 N
D) 233 N

Accepted Answer

In static equilibrium, the vertical component of the tension force in the rope must balance the weight of the trapeze artist. 
Using trigonometry, we can find that the vertical component of the tension force equals T * cos(30°), where T is the tension force in the rope. 
Thus, we have: 
T * cos(30°) = 700 N 
T = 700 N / cos(30°) 
T ≈ 808 N

Therefore, the best choice is C, because it is the closest to the calculated value of the tension force.

Question 10

A 25-g traffic light hangs midway on a cable between two poles 40 meters apart.If the sag in the cable is 0.40 meters,what is the tension in each side of the cable?
A) 12,200 N
B) 9800 N
C) 4900 N
D) 6100 N

Accepted Answer

The answer of A 25-g traffic light hangs midway on...

Question 11

A 100-kg box is placed on a ramp.As one end of the ramp is raised,the box begins to move downward just as the angle of inclination reaches 25°.What is the coefficient of static friction between box and ramp?
A) 0.15
B) 0.27
C) 0.47
D) 0.95

Accepted Answer

The force due to gravity acting on the box is given by Fg = mg, where m is the mass of the box and g is the acceleration due to gravity. 
Fg = (100 kg)(9.81 m/s^2) = 981 N 
The force acting parallel to the ramp is given by Fpara = Fg sinθ, where θ is the angle of inclination of the ramp. 
Fpara = (981 N)sin(25°) = 416 N 
The force of static friction acting opposite to the direction of motion is given by Fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force acting perpendicular to the surface. 
N = Fg cosθ = (981 N)cos(25°) = 889 N 
Therefore, 
Fs ≤ μsN 
Fs = Fpara = 416 N 
Substituting for Fs, 
μsN = 416 N 
μs(889 N) = 416 N 
μs = 0.468
The closest answer choice to this is C (0.47), so the correct answer is C.

Question 12

A 200-N block,on a 30° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.60.The force due to friction is:
A) 104 N.
B) 100 N.
C) 60 N.
D) 0 N.

Accepted Answer

The force due to friction is given by the product of the coefficient of static friction and the normal force acting on the block. The normal force is equal to the component of the weight of the block perpendicular to the plane, which is given by:

N = mg cosθ = 200 N × cos30° ≈ 173.2 N

Therefore, the force due to friction is:

f_s = μ_s N = 0.60 × 173.2 N ≈ 103.9 N ≈ 100 N (to one significant figure)

Since the block is being held motionless by friction, the force due to friction must be equal and opposite to the component of the weight of the block parallel to the plane:

f_s = mg sinθ = 200 N × sin30° ≈ 100 N

Therefore, the correct answer is B) 100 N.

Question 13

The coefficient of static friction between the tires of a car and the street is µ_s = 0.58.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked)without slipping? A) 22.5° B) 30° C) 38° D) 45°

Accepted Answer

The answer of The coefficient of static friction between the...

Question 14

A sled weighs 100 N.It is held in place on a frictionless 30° slope by a rope attached to a stake at the top;the rope is parallel to the slope.What is the normal force acting on the sled?
A) 94 N
B) 87 N
C) 50 N
D) 34 N

Accepted Answer

The normal force is perpendicular to the surface and acts to counteract the weight of the object. Since there is no friction, the force of the rope only acts to keep the sled in place and does not affect the normal force. To find the normal force, we need to resolve the weight force into two components: one parallel to the slope and one perpendicular. 
The component of the weight force perpendicular to the slope is given by: 
F(perpendicular) = W * cos(θ) 
= 100 N * cos(30°) 
= 86.6 N

Therefore, the normal force is equal and opposite to F(perpendicular), and its value is approximately 87 N (closest to B).

Question 15

A sled is on a horizontal surface and a rope with tension T is pulling on it at an angle θ above the horizontal.The sled had mass M and does not move due to the frictional force.In this case the coefficient of friction is μ_s.In terms of the variables given and g,what is the magnitude of the frictional force in this case? A) μ_sMg B) Tcosθ C) Tsinθ D) μ_sMgcosθ

Accepted Answer

The answer of A sled is on a horizontal surface...

Question 16

A 500-N tightrope walker stands at the center of the rope such that each half of the rope makes an angle of 15.0° with the horizontal.What is the tension in the rope?
A) 1440 N
B) 970 N
C) 480 N
D) 2900 N

Accepted Answer

First, we can draw a free-body diagram of the tightrope walker. The only forces acting on the tightrope walker are the weight force (500 N) acting downward and the tension force in the rope acting upward.

Next, we can break up the tension force into its horizontal and vertical components. Each half of the rope makes an angle of 15.0° with the horizontal, so the angle between the tension force and the horizontal is also 15.0°. Therefore, the vertical component of the tension force is T sin 15° and the horizontal component is T cos 15°.

Since the tightrope walker is in equilibrium, the net force in both the x and y directions must be zero. Therefore, the vertical components of the tension forces on each side of the tightrope walker must balance out the weight force:

T sin 15° + T sin 15° = 500 N
2T sin 15° = 500 N
T = 500 N / (2 sin 15°) ≈ 970 N

Therefore, the tension in the rope is approximately 970 N, which is answer choice B.

Question 17

A helicopter is traveling at 54 m/s at a constant altitude of 100 m over a level field.If a wheel falls off the helicopter,with what speed will it hit the ground? Note: g = 9.8 m/s² and air resistance is negligible. A) 40 m/s B) 50 m/s C) 60 m/s D) 70 m/s

Accepted Answer

The answer of A helicopter is traveling at 54 m/s...

Question 18

A projectile is fired over a horizontal surface at an angle of elevation of 70°.Neglecting air resistance,what are possible angles in flight between the acceleration vector and the velocity vector?
A) 170° and 40°
B) 120° and 70°
C) 90° and 15°
D) none of the above

Accepted Answer

The answer of A projectile is fired over a horizontal...

Question 19

John throws a baseball from the outfield from shoulder height,at an initial velocity of 39.2 m/s at an initial angle of 30.0° with respect to the horizontal.The ball is in its trajectory for a total interval of 4.00 s before the third baseman catches it at an equal shoulder-height level.(Assume air resistance is negligible. )What is the ball's horizontal displacement?
A) 76.4 m
B) 38.2 m
C) 136 m
D) zero

Accepted Answer

The answer of John throws a baseball from the outfield...

Question 20

A 4000-N weight is held suspended in equilibrium by two cables.Cable 1 applies a horizontal force to the right of the object and has a tension T₁.Cable 2 applies a force upward and to the left at an angle of 37.0° to the negative x axis and has a tension T₂.What is the tension T₁? A) 4000 N B) 6640 N C) 6250 N D) 5310 N

Accepted Answer

The answer of A 4000-N weight is held suspended in...

Quiz 4: The Laws of Motion: Part A

A ball is rolled horizontally off a table with an initial speed of 0.18 m/s.A stopwatch measures the ball's trajectory time from table to the floor to be 0.30 s.What is the height of the table? Note: g = 9.8 m/s² and air resistance is negligible.

A 4000-N weight is suspended in equilibrium by two cables.Cable 1 applies a horizontal force to the right of the object and has a tension T₁.Cable 2 applies a force upward and to the left at an angle of 37.0° to the negative x axis and has a tension T₂.Find T_2.

A baseball thrown from the outfield is released from shoulder height at an initial velocity of 39.2 m/s at an initial angle of 30.0° with respect to the horizontal.What is the maximum vertical displacement that the ball reaches during its trajectory?

A trapeze artist,with swing,weighs 700 N;he is momentarily held to one side by his partner so that the swing ropes make an angle of 30.0° with the vertical.In such a condition of static equilibrium,what is the horizontal force being applied by the partner?

A 200-N crate rests on a ramp;the maximum angle just before it slips is 21° with the horizontal.What is the coefficient of static friction between crate and ramp surfaces?

A 500-N tightrope walker stands at the center of the rope.If the rope can withstand a tension of 1400 N without breaking,what is the minimum angle the rope can make with the horizontal?

A ball is rolled horizontally off a table with an initial speed of 0.18 m/s.A stopwatch measures the ball's trajectory time from table to the floor to be 0.30 s.How far away from the table does the ball land? Note: g = 9.8 m/s² and air resistance is negligible.

A trapeze artist,with swing,weighs 700 N;he is being held to one side by his partner so that the swing ropes make an angle of 30.0° with the vertical.In such a condition of static equilibrium what is the tension in the rope?

A 25-g traffic light hangs midway on a cable between two poles 40 meters apart.If the sag in the cable is 0.40 meters,what is the tension in each side of the cable?

A 100-kg box is placed on a ramp.As one end of the ramp is raised,the box begins to move downward just as the angle of inclination reaches 25°.What is the coefficient of static friction between box and ramp?

A 200-N block,on a 30° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.60.The force due to friction is:

The coefficient of static friction between the tires of a car and the street is µ_s = 0.58.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping?

A sled weighs 100 N.It is held in place on a frictionless 30° slope by a rope attached to a stake at the top;the rope is parallel to the slope.What is the normal force acting on the sled?

A 500-N tightrope walker stands at the center of the rope such that each half of the rope makes an angle of 15.0° with the horizontal.What is the tension in the rope?

A helicopter is traveling at 54 m/s at a constant altitude of 100 m over a level field.If a wheel falls off the helicopter,with what speed will it hit the ground? Note: g = 9.8 m/s² and air resistance is negligible.

A projectile is fired over a horizontal surface at an angle of elevation of 70°.Neglecting air resistance,what are possible angles in flight between the acceleration vector and the velocity vector?

A 4000-N weight is held suspended in equilibrium by two cables.Cable 1 applies a horizontal force to the right of the object and has a tension T₁.Cable 2 applies a force upward and to the left at an angle of 37.0° to the negative x axis and has a tension T₂.What is the tension T₁?