A force of 15 N is applied at an angle of 30 degrees with the horizontal on a 0.75 block at rest on a frictionless surface.
a. What is the magnitude of the resulting acceleration of the block?
b. What is the magnitude of the normal force?
Is the mass of the block 0.75 kg? Do not omit the dimensions that go with your numbers.
The force component in the direction of motion is
Fx = 15 cos30 = 13.0 N
Since there is no opposing friction,
a = Fx/M = 0.37 m/s^2
a. Well, well, well, looks like we have a force playing a game of tug-of-war with a block! But don't worry, I've got the answer for you! To find the magnitude of acceleration, we can use Newton's second law, which states that force is equal to mass times acceleration. Since the block has a mass of 0.75 kg, we can calculate the acceleration with the formula: F = m * a. Rearranging the equation, we have a = F / m. Plugging in the numbers, we get a = 15 N / 0.75 kg. Crunching it down, the magnitude of the resulting acceleration is 20 m/s². Time to get moving!
b. Ah, the normal force, the unsung hero of surfaces everywhere. Well, my friend, on a frictionless surface, the normal force is equal to the weight of the block. To find the magnitude of the normal force, we can simply calculate the weight of the block. The weight is given by the formula: weight = mass * gravity. Plugging in the values, we get weight = 0.75 kg * 9.8 m/s². Crunching the numbers, the magnitude of the normal force is roughly 7.35 N. It's always nice to have a balanced support system in life, isn't it?
To solve this problem, we'll need to break down the force into its horizontal and vertical components.
a. Magnitude of the resulting acceleration:
The horizontal component of the force (F_hor) can be found using trigonometry, specifically using cosine:
F_hor = F * cos(theta)
where F is the magnitude of the force (15 N) and theta is the angle (30 degrees).
Plugging in the values:
F_hor = 15 N * cos(30 degrees)
F_hor ≈ 15 N * 0.866
F_hor ≈ 12.99 N
Since there is no friction and the block is initially at rest, the horizontal force (F_hor) is equal to the net force (F_net). Newton's second law states that the net force on an object is equal to its mass (m) multiplied by its acceleration (a):
F_net = m * a
Therefore, we can rewrite the equation as:
F_net = F_hor = m * a
Substituting in the values we found for F_hor and the mass of the block (0.75 kg):
12.99 N = 0.75 kg * a
a = 12.99 N / 0.75 kg
a ≈ 17.32 m/s²
Therefore, the magnitude of the resulting acceleration of the block is approximately 17.32 m/s².
b. Magnitude of the normal force:
The vertical component of the force (F_ver) can be found using trigonometry, specifically using sine:
F_ver = F * sin(theta)
where F is the magnitude of the force (15 N) and theta is the angle (30 degrees).
Plugging in the values:
F_ver = 15 N * sin(30 degrees)
F_ver ≈ 15 N * 0.5
F_ver ≈ 7.5 N
The normal force (N) is equal in magnitude and opposite in direction to the force that gravity exerts perpendicular to the surface of the block. Since the force component in the vertical direction is greater than the force of gravity, the normal force will also be greater than the weight of the block.
Therefore, the magnitude of the normal force is approximately 7.5 N.
To find the magnitude of the resulting acceleration of the block, we need to use Newton's second law of motion. The formula for Newton's second law is:
F = ma
where F is the net force acting on the object, m is the mass of the object, and a is the acceleration of the object.
In this case, the force applied is 15 N and it is at an angle of 30 degrees with the horizontal. We can split this force into its horizontal and vertical components using trigonometry.
The horizontal component of the force can be found using the formula:
F_horizontal = F * cos(theta)
where theta is the angle between the applied force and the horizontal. In this case, F_horizontal = 15 N * cos(30) ≈ 12.99 N.
Since the block is on a frictionless surface, there is no horizontal force opposing the motion. Therefore, the net horizontal force is equal to the horizontal component of the applied force:
F_net_horizontal = F_horizontal = 12.99 N
Using Newton's second law, we can find the magnitude of the resulting acceleration of the block:
F_net_horizontal = ma
12.99 N = 0.75 kg * a
Solving for a, we get:
a ≈ 12.99 N / 0.75 kg ≈ 17.32 m/s^2
Therefore, the magnitude of the resulting acceleration of the block is approximately 17.32 m/s^2.
To find the magnitude of the normal force, we need to consider the forces acting on the block in the vertical direction. The only vertical force acting on the block is the force of gravity.
The formula for the force of gravity is:
F_gravity = mg
where m is the mass of the object (0.75 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 0.75 kg * 9.8 m/s^2 ≈ 7.35 N
Since there is no vertical acceleration (the block is not moving vertically), the magnitude of the normal force is equal to the force of gravity:
Magnitude of normal force = F_gravity = 7.35 N
Therefore, the magnitude of the normal force is approximately 7.35 N.