Question 1

FIGURE 4-11 -A 4.00-kg block slides down a frictionless inclined plane with an acceleration 3.00 m/s². What is the angle of the incline above horizontal? A)35.3° B)45.2° C)17.8° D)23.6° E)53.7°

Accepted Answer

Using the formula $a=g\sin	heta$ where $a$ is the acceleration, $g$ is the acceleration due to gravity, and $	heta$ is the angle of the incline, we can solve for $	heta$:
$$3.00	ext{ m/s}^2=g\sin	heta$$
$$\sin	heta=\frac{3.00	ext{ m/s}^2}{9.81	ext{ m/s}^2}=0.306$$
Taking the inverse sine, we get $	heta=\boxed{17.8^\circ	ext{ (C)}}$.

Question 2

FIGURE 4-13 
-A traffic light of weight 100 N is supported by two ropes as shown in Fig. 4-13. What are the tensions in the ropes?
A)50 N
B)56 N
C)63 N
D)66 N
E)83 N

Accepted Answer

The answer of FIGURE 4-13 
-A traffic light of weight...

Question 3

FIGURE 4-12 -A 60.0-kg person rides in elevator while standing on a scale. The elevator has an acceleration 2.00 m/s² upward. What is the reading on the scale? A)589 N B)709 N C)469 N D)zero E)349 N

Accepted Answer

We can start by drawing a free body diagram of the person. There are two forces acting on the person: their weight (mg) downwards and the normal force (N) upwards from the scale. 
Using Newton's second law, we can write the equation: 
N - mg = ma 
where a is the acceleration of the elevator. 
Plugging in the given values: 
N - (60.0 kg)(9.81 m/s^2) = (60.0 kg)(2.00 m/s^2) 
Solving for N: 
N = (60.0 kg)(2.00 m/s^2) + (60.0 kg)(9.81 m/s^2) 
N = 1182 N 
However, this is the apparent weight, which includes the force from the acceleration. To find the actual reading on the scale (which is the normal force), we need to subtract the force from the acceleration: 
N(actual) = N(apparent) - ma 
N(actual) = 1182 N - (60.0 kg)(2.00 m/s^2) 
N(actual) = 709 N 
Therefore, the answer is B) 709 N.

Question 4

FIGURE 4-12 
-A 6.00-kg block is in contact with a 4.00-kg block on a frictionless surface as shown in Fig. 4-12. The 6.00-kg block is being pushed by a 20.0-N force toward the 4.00-kg block. What is the magnitude of the force of the 6.00-kg block on the 4.00-kg block?
A)6.00 N
B)12.0 N
C)8.00 N
D)4.00 N
E)10.0 N

Accepted Answer

According to Newton's third law, the force of the 6.00-kg block on the 4.00-kg block is equal in magnitude but opposite in direction to the force of the 4.00-kg block on the 6.00-kg block. The force of the 4.00-kg block on the 6.00-kg block is 8.00 N, so the force of the 6.00-kg block on the 4.00-kg block is also 8.00 N.

Question 5

FIGURE 4-13 
-An object is on a frictionless inclined plane. The plane is inclined at an angle of 30° with the horizontal. What is the object's acceleration?
A)0.50 g
B)0.56 g
C)0.68 g
D)0.87 g
E)1.0 g

Accepted Answer

The acceleration of an object on a frictionless inclined plane is given by g*sin(θ). For θ = 30°, sin(30°) = 0.5, so the acceleration is 0.5g.

Question 6

FIGURE 4-12 -A 60.0-kg person rides in an elevator while standing on a scale. The scale reads 400 N. What is the acceleration of the elevator? A)3.14 m/s² downward B)6.67 m/s² downward C)zero D)9.81 m/s² downward E)6.67 m/s² upward

Accepted Answer

The scale reading is less than the person's weight, so the elevator must be accelerating downward. The acceleration is:a = (F - mg) / m = (400 N - 60.0 kg * 9.81 m/s^2) / 60.0 kg = 3.14 m/s^2 downward

Quiz 4: Dynamics: Newtons Laws of Motion

FIGURE 4-11 -A 4.00-kg block slides down a frictionless inclined plane with an acceleration 3.00 m/s2. What is the angle of the incline above horizontal?

FIGURE 4-13 -A traffic light of weight 100 N is supported by two ropes as shown in Fig. 4-13. What are the tensions in the ropes?

FIGURE 4-12 -A 60.0-kg person rides in elevator while standing on a scale. The elevator has an acceleration 2.00 m/s2 upward. What is the reading on the scale?

FIGURE 4-12 -A 6.00-kg block is in contact with a 4.00-kg block on a frictionless surface as shown in Fig. 4-12. The 6.00-kg block is being pushed by a 20.0-N force toward the 4.00-kg block. What is the magnitude of the force of the 6.00-kg block on the 4.00-kg block?

FIGURE 4-13 -An object is on a frictionless inclined plane. The plane is inclined at an angle of 30° with the horizontal. What is the object's acceleration?

FIGURE 4-12 -A 60.0-kg person rides in an elevator while standing on a scale. The scale reads 400 N. What is the acceleration of the elevator?

FIGURE 4-11 -A 4.00-kg block slides down a frictionless inclined plane with an acceleration 3.00 m/s². What is the angle of the incline above horizontal?

FIGURE 4-12 -A 60.0-kg person rides in elevator while standing on a scale. The elevator has an acceleration 2.00 m/s² upward. What is the reading on the scale?