1)When a 10-kg crate is pushed across a frictionless horizontal floor with a force of 14 N, directed 25° below the horizontal, the magnitude of the normal force of the floor on the crate is (g = 10 m/s2):
2)A constant force of 2.8 N is exerted for 9.0 s on a 6.1-kg object initially at rest. The change in speed of this object will be:
3)When a certain force is applied to the 1-kg standard mass its acceleration is 5.3 m/s2. When the same force is applied to another object its acceleration is one-fifth as much. The mass of the object is:
4) Two forces are applied to a 7.6-kg crate; one is 8.5 N to the north and the other is 7.8 N to the west. The magnitude of the acceleration of the crate is:
pls help me
1) To find the magnitude of the normal force of the floor on the crate, we need to analyze the forces acting on the crate.
The only force acting horizontally is the push force, which has a magnitude of 14 N and is directed 25° below the horizontal.
We can resolve this force into its horizontal and vertical components using trigonometry. The horizontal component is given by 14 N * cos(25°) = 12.68 N.
Since the floor is frictionless, the normal force is equal to the vertical component of the push force. In this case, the vertical component is given by 14 N * sin(25°) = 6.01 N.
Therefore, the magnitude of the normal force of the floor on the crate is 6.01 N.
2) The change in speed of an object is given by the equation:
Δv = (F/m) * Δt
Where Δv is the change in velocity (speed), F is the applied force, m is the mass of the object, and Δt is the time interval.
In this case, the force is 2.8 N, the mass is 6.1 kg, and the time interval is 9.0 s. Plugging these values into the equation, we get:
Δv = (2.8 N / 6.1 kg) * 9.0 s = 4.1049 m/s
Therefore, the change in speed of the object will be approximately 4.10 m/s.
3) The relationship between force, mass, and acceleration is given by Newton's second law of motion as F = m * a.
In this scenario, we know that when the standard mass of 1 kg is subjected to a force, its acceleration is 5.3 m/s².
Let's consider the other object with an unknown mass. When the same force is applied to it, its acceleration is one-fifth (1/5) as much.
So, we can set up the following equation:
(1 kg * 5.3 m/s²) = (m kg * (1/5) * 5.3 m/s²)
Simplifying the equation, we get:
5.3 m/s² = (m/5) * 5.3 m/s²
5.3 m/s² = m m/s²
m = 5.3 kg
Therefore, the mass of the object is 5.3 kg.
4) To find the magnitude of the acceleration of the crate, we need to use Newton's second law of motion, F = m * a, where F is the net force applied to the object, m is the mass of the object, and a is the acceleration.
In this scenario, the two forces applied to the crate are 8.5 N to the north and 7.8 N to the west. These forces act in perpendicular directions, so we can use the Pythagorean theorem to find the net force:
Net force = √(8.5 N)^2 + (7.8 N)^2
Net force = √(72.25 N^2 + 60.84 N^2)
Net force = √133.09 N^2
Net force = 11.53 N
To find the acceleration of the crate, we can use the formula:
a = (ΣF) / m
Where ΣF is the net force and m is the mass of the crate, given as 7.6 kg.
a = 11.53 N / 7.6 kg
a = 1.52 m/s^2
Therefore, the magnitude of the acceleration of the crate is 1.52 m/s².
1) To find the magnitude of the normal force, we need to consider the forces acting on the crate. In this case, the only force acting on the crate is a horizontal force of 14 N directed 25° below the horizontal. Since the floor is frictionless, there is no force opposing the motion of the crate.
To find the normal force, we can use the horizontal component of the applied force. The horizontal component can be calculated using the formula: F_horizontal = F_applied * cos(angle), where F_applied is the applied force and angle is the angle between the applied force and the horizontal direction.
So, F_horizontal = 14 N * cos(25°) = 12.636 N (approximately)
The magnitude of the normal force is equal to the weight of the crate, which can be calculated using the formula: weight = mass * acceleration due to gravity, where mass is the mass of the crate and acceleration due to gravity (g) is 10 m/s².
weight = 10 kg * 10 m/s² = 100 N
Since the crate is on a horizontal floor, the magnitude of the normal force is equal to the weight of the crate. Therefore, the magnitude of the normal force is 100 N.
2) To calculate the change in speed, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. The net force is given by the constant force of 2.8 N, and the mass of the object is 6.1 kg.
So, we can calculate the acceleration using the formula: acceleration = net force / mass.
acceleration = 2.8 N / 6.1 kg ≈ 0.459 m/s²
With the acceleration known, we can use the equation of motion: final speed² = initial speed² + 2 * acceleration * displacement, where the initial speed is 0 m/s (as the object is initially at rest), and the displacement is calculated using the equation: displacement = acceleration * time² / 2.
displacement = 0.459 m/s² * (9.0 s)² / 2 ≈ 18.42 m
Substituting these values into the equation of motion, we can solve for the final speed:
final speed² = 0 + 2 * 0.459 m/s² * 18.42 m ≈ 16.90 m²/s²
Taking the square root of both sides, we find:
final speed ≈ √(16.90 m²/s²) ≈ 4.11 m/s
Therefore, the change in speed of the object will be approximately 4.11 m/s.
3) To solve this problem, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration.
We are given that the force applied to the 1-kg standard mass produces an acceleration of 5.3 m/s². So, we can write the equation as: force = mass * acceleration.
Force = 1 kg * 5.3 m/s² = 5.3 N
Now, we are told that the same force is applied to another object (mass unknown), resulting in an acceleration one-fifth as much as the standard mass. Let's denote the mass of the object as m.
Force = m * (5.3 m/s² / 5) = 5.3 N
Simplifying the equation, we find:
m * 1.06 m/s² = 5.3 N
m ≈ 5 N / 1.06 m/s² ≈ 4.72 kg
Therefore, the mass of the object is approximately 4.72 kg.
4) To find the acceleration of the crate, we need to determine the net force acting on it. The net force is the vector sum of the individual forces.
The two forces acting on the crate are 8.5 N to the north and 7.8 N to the west. Since forces are vectors, we need to find the resultant vector by using vector addition.
Let's denote the north direction as positive (upwards) and the west direction as negative (to the left).
The northward force of 8.5 N and the westward force of 7.8 N can be represented as vectors: F₁ = 8.5 N (north) and F₂ = 7.8 N (west).
Since the forces are at right angles to each other, we can use the Pythagorean theorem to find the magnitude of the resultant force (F) applied to the crate:
F = √(F₁² + F₂²)
F = √((8.5 N)² + (7.8 N)²)
F ≈ √(72.25 N² + 60.84 N²)
F ≈ √(133.09 N²)
F ≈ 11.53 N (approximately)
To find the acceleration, we can use Newton's second law of motion: net force = mass * acceleration. The net force applied to the crate is 11.53 N, and the mass of the crate is 7.6 kg.
So, we can rearrange the equation to solve for acceleration (a):
acceleration = net force / mass
acceleration = 11.53 N / 7.6 kg
acceleration ≈ 1.52 m/s²
Therefore, the magnitude of the acceleration of the crate is approximately 1.52 m/s².
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