Lesson 3 : Newton's Second Law Of Motion

Answer: 3 m/s/s

The original value of 2 m/s/s must be multiplied by 3 (since a and F are directly proportional) and divided by 2 (since a and m are inversely proportional)

Answer: 12 m/s/s

The original value of 2 m/s/s must be multiplied by 3 (since a and F are directly proportional) and divided by 1/2 (since a and m are inversely proportional)

Use Fnet= m * a with Fnet = 15 N and a = 5 m/s/s.

So (15 N) = (m)*(5 m/s/s)

And m = 3.0 kg

The explanation for the given correct answer is that determining the value of individual forces acting upon an object indeed involves applying Newton's second law (Fnet=m*a) and understanding the concept of net force. Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. By considering the net force and applying this equation, one can determine the value of individual forces acting on the object. Therefore, the statement is true.

The statement is true because according to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration. Therefore, if the values of mass and acceleration are known, the net force can be determined using the equation Fnet = m * a.

According to Newton's first law of motion, an object at rest will stay at rest, and an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Therefore, an object will only accelerate if there is a net or unbalanced force acting upon it. This means that the presence of an unbalanced force will cause a change in the object's speed, direction, or both, confirming that the statement is true.

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A free-body diagram is a visual representation of an object showing all the forces acting on it. By constructing a free-body diagram, we can identify and calculate the individual forces acting on the object. This helps us analyze the motion and equilibrium of the object. Therefore, the statement that a free-body diagram is constructed and the individual forces acting upon the object are identified and calculated is true.

Objects that are undergoing free fall are not encountering a significant force of air resistance. This means that the only force acting on these objects is gravity, causing them to fall towards the ground. Therefore, the statement is true.

The explanation for the given correct answer is that Newton's second law of motion (Fnet = m*a) can be applied to analyze the motion of objects falling under the influence of gravity. This law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In the case of objects falling under gravity, the force of gravity is the net force acting on the object, and the acceleration is the acceleration due to gravity. Therefore, Newton's second law can be used to analyze the motion of objects falling under gravity. Additionally, the statement mentions that this law can also be applied to objects falling under the dual influence of gravity and air resistance.

Cognitive scientists have found that physics students often hold onto certain beliefs about motion, known as misconceptions, even when presented with evidence that contradicts these beliefs. These misconceptions can impede their ability to learn and understand physics concepts. Therefore, the statement that cognitive scientists have shown that physics students come into class with these beliefs and that they hinder further learning is true.

The statement is true because the ratio Fnet/m, which represents the force experienced by an object per unit mass, is indeed referred to as the gravitational field strength. On Earth's surface, this value is approximately 9.8 N/kg.

Fgrav = 7.50 N; Fnorm = 7.50 N; Ffrict = 3.07 N; Fnet = 1.18 N, right; a = 1.54 m/s/s, right

The starting point for any problem such as this is the construction of a free-body diagram in which you show all the individual forces which are acting upon the book. There are two vertical forces - gravity and normal force. There are two horizontal forces - friction and the applied force.
Since there is no vertical acceleration, normal force = gravity force. Each of these forces can be determined using the equation Fgrav = m • g = (0.765 kg) • (9.8 m/s/s) = 7.497 N

The force of friction can be determined using the equation Ffrict = mu • Fnorm. So Ffrict = (0.410) • (7.497 N) = (3.0737 ... N)

The Fnet is the vector sum of all the forces: 4.25 N, right plus 3.0737 ... N, left = 1.176... N, right.

Finally, a = Fnet / m = (1.176... N) / (0.765 kg) = 1.54 m/s/s.

Fgrav = 4.90 N; Fnorm = 4.90 N; Fnet = 1.73 N, right; a = 3.46 m/s/s, right

The starting point for any problem such as this is the construction of a free-body diagram in which you show all the individual forces which are acting upon the book. There are two vertical forces - gravity and normal force. There are two horizontal forces - friction and the applied force.
Since there is no vertical acceleration, normal force = gravity force. Each of these forces could be determined using the equation Fgrav = m • g = (0.500 kg)•(9.8 m/s/s) = 4.90 N.

The Fnet is the vector sum of all the forces: 2.45 N, right plus 0.72 N, left = 1.73 N, right.

Finally, a = Fnet / m = (1.73. N) / (0.500 kg) = 3.46 m/s/s.

A 3-kg object experiences an acceleration of 4 m/s/s.

Ffrict = 28.7 N, left

The starting point for any problem such as this is the construction of a free-body diagram in which you show all the individual forces which are acting upon the book. There are two vertical forces - gravity and normal force. There are two horizontal forces - friction and the applied force.
Since there is no vertical acceleration, normal force = gravity force. Each of these forces can be determined using the equation Fgrav = m • g = (4.525 kg) • (9.8 m/s/s) = 44.296 N

The net force can be determined from knowledge of the mass and acceleration of the sled. Fnet = m • a = (4.52 kg) • (1.28 m/s/s) = 5.7856 N, right.

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces. So 5.7856 N, right = 34.5 N, right + Ffrict. Therefore, Ffrict = 28.7 N, left.

a = 4.46 m/s2

The solution here will use the approach of a system analysis and an individual object analysis. The free-body diagrams for the system and for the car are shown below.



For the system: Fnet = 2.50x104 N and msystem = 5.60x103 kg. So

a = Fnet/m = (2.50x104 N) / (5.60x103 kg) = 4.4643 m/s2

a = 1.32 m/s2

The solution here will use the approach of a system analysis and an individual object analysis . The free-body diagrams for the system and for the 3-kg object are shown below.


For the system: Ffrict = μ•Fnorm = 0.120• 98.0 N = 11.76 N

Fnet = 25.0 N - 11.76 N = 13.24 N and msystem = 10.0 kg.

So a = Fnet/m = (13.24 N) / (10.0 kg) = 1.324 m/s2 (round to 1.32 m/s2)

Fgrav = 45.9 N; Fnorm = 45.9 N; Ffrict = 5.32 N; Fnet = 3.81 N, right; mu = 0.116


The starting point for any problem such as this is the construction of a free-body diagram in which you show all the individual forces which are acting upon the book. There are two vertical forces - gravity and normal force. There are two horizontal forces - friction and the applied force.

Since there is no vertical acceleration, normal force = gravity force. Each of these forces can be determined using the equation Fgrav = m • g = (4.68 kg) • (9.8 m/s/s) = 45.864 N

The net force can be determined from knowledge of the mass and acceleration of the sled. Fnet = m • a = (4.68 kg) • (0.815 m/s/s) = 3.8142 N, right.

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces. So 3.81 N, right = 9.13 N, right + Ffrict. Therefore, Ffrict = 5.32 N, left.

The coefficient of friction can now be determined as the ratio of friction force to normal force. "mu" = Ffrict / Fnorm = (5.32 N) / (45.864 N) = 0.116.

A 6-kg object experiences an acceleration of 2 m/s/s.