A force of 200 N is being exerted on a blocki, which has a mass of 10 kg and is on 30 degrees inclined plane. The force is acting up the plane and parallel to it. The coefficient of friction between the block and the plane is .3 and the plane is 12 m long.

a. What is the weight of the block
b. What is the parallel force acting on the block down the incline?
c. What is the normal force acting on the block?
d. What is the frictional force acting on the block?
e. What is the net force acting on the block?
f. What is the acceleration of the block?
g. If the block starts in the middle of the plane, what is its speed at the end of the incline?

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To answer these questions, we need to break down the problem into several steps and calculate each variable accordingly.

a. What is the weight of the block?

The weight of an object can be calculated using the formula:

Weight = Mass × Acceleration due to gravity

Given that the mass of the block is 10 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight:

Weight = 10 kg × 9.8 m/s^2 = 98 N

Therefore, the weight of the block is 98 N.

b. What is the parallel force acting on the block down the incline?

The parallel force acting on the block down the incline is the component of the force that acts parallel to the incline. We can calculate this force using the formula:

Parallel Force = Force × sin(θ)

Given that the force acting on the block is 200 N and the angle of the incline is 30 degrees, we can calculate the parallel force:

Parallel Force = 200 N × sin(30 degrees) = 100 N

Therefore, the parallel force acting on the block down the incline is 100 N.

c. What is the normal force acting on the block?

The normal force acting on an object on an inclined plane is the perpendicular force exerted by the plane on the block. It can be calculated using the formula:

Normal Force = Weight × cos(θ)

Given that the weight of the block is 98 N and the angle of the incline is 30 degrees, we can calculate the normal force:

Normal Force = 98 N × cos(30 degrees) = 84.85 N (approximately)

Therefore, the normal force acting on the block is approximately 84.85 N.

d. What is the frictional force acting on the block?

The frictional force can be calculated using the formula:

Frictional Force = Coefficient of Friction × Normal Force

Given that the coefficient of friction between the block and the plane is 0.3 and the normal force is approximately 84.85 N, we can calculate the frictional force:

Frictional Force = 0.3 × 84.85 N = 25.45 N (approximately)

Therefore, the frictional force acting on the block is approximately 25.45 N.

e. What is the net force acting on the block?

To calculate the net force, we need to consider the forces acting parallel and perpendicular to the incline.

Net Force = Parallel Force - Frictional Force - Weight × sin(θ)

Given that the parallel force is 100 N, the frictional force is approximately 25.45 N, the weight is 98 N, and the angle of the incline is 30 degrees, we can calculate the net force:

Net Force = 100 N - 25.45 N - 98 N × sin(30 degrees)
Net Force = 100 N - 25.45 N - 98 N × 0.5
Net Force = 100 N - 25.45 N - 49 N
Net Force = 25.55 N

Therefore, the net force acting on the block is 25.55 N.

f. What is the acceleration of the block?

To calculate the acceleration of the block, we can use Newton's second law of motion:

Acceleration = Net Force / Mass

Given that the net force is 25.55 N and the mass of the block is 10 kg, we can calculate the acceleration:

Acceleration = 25.55 N / 10 kg = 2.55 m/s^2

Therefore, the acceleration of the block is 2.55 m/s^2.

g. If the block starts in the middle of the plane, what is its speed at the end of the incline?

To calculate the speed of the block at the end of the incline, we can use the equations of motion. Since the block starts from rest, we can use the following equation:

v^2 = u^2 + 2as

Where:
v = final velocity (unknown)
u = initial velocity (0 m/s since it starts from rest)
a = acceleration (2.55 m/s^2)
s = distance traveled (12 m, the length of the incline)

Using the equation, we can solve for the final velocity:

v^2 = 0^2 + 2 × 2.55 m/s^2 × 12 m
v^2 = 0 + 61.2 m^2/s^2
v = √(61.2) m/s
v ≈ 7.82 m/s

Therefore, the speed of the block at the end of the incline is approximately 7.82 m/s.