A 120-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 450 N. For the first 11 m the floor is frictionless, and for the next 11 m the coefficient of friction is 0.30. What is the final speed of the crate?

Calculate the work done during each of the two periods (frictionless and with friction) and add them.

Equate the Total Work sum to the final kinetic energy and solve for V. The M's will cancel out.

To calculate the final speed of the crate, we first need to determine the acceleration of the crate in each segment and then use that to find the overall acceleration.

First, let's calculate the acceleration in the first 11 m where the floor is frictionless. We can 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.

Given:
Mass of the crate (m): 120 kg
Net force (F): 450 N (since there is no friction in the first 11 m)

We can rearrange the equation to solve for acceleration (a):
a = F / m
a = 450 N / 120 kg
a ≈ 3.75 m/s²

The acceleration in the first 11 m where the floor is frictionless is approximately 3.75 m/s².

Next, let's calculate the acceleration in the next 11 m where the coefficient of friction is 0.30. The force of friction can be calculated using the equation:

Force of friction (Ff) = coefficient of friction (μ) * normal force (Fn)
Normal force (Fn) = mass (m) * acceleration due to gravity (g)

Given:
Mass of the crate (m): 120 kg
Coefficient of friction (μ): 0.30
Acceleration due to gravity (g): 9.8 m/s²

Fn = m * g
Fn = 120 kg * 9.8 m/s²
Fn ≈ 1176 N

Ff = μ * Fn
Ff = 0.30 * 1176 N
Ff ≈ 352.8 N

The force of friction in the next 11 m is approximately 352.8 N.

Now, let's find the acceleration in the next 11 m using Newton's second law of motion:

Net force (F) = applied force (450 N) - force of friction (352.8 N)

a = F / m
a = (450 N - 352.8 N) / 120 kg
a ≈ 0.81 m/s²

The acceleration in the next 11 m where the coefficient of friction is 0.30 is approximately 0.81 m/s².

Finally, to find the overall acceleration, we can take the average of the two accelerations:

Overall acceleration (a) = (3.75 m/s² + 0.81 m/s²) / 2
a ≈ 2.28 m/s²

The overall acceleration is approximately 2.28 m/s².

Now, we can use the equation of motion to find the final speed of the crate. The equation relating final speed (v), initial speed (u), acceleration (a), and distance (s) is:

v² = u² + 2as

Given:
Initial speed (u) = 0 (since the crate starts from rest)
Distance (s) = 11 m + 11 m = 22 m
Acceleration (a) ≈ 2.28 m/s²

v² = 0² + 2 * 2.28 m/s² * 22 m
v² ≈ 101.28 m²/s²

Taking the square root of both sides:

v ≈ √(101.28 m²/s²)
v ≈ 10.06 m/s

Therefore, the final speed of the crate is approximately 10.06 m/s.

To find the final speed of the crate, we need to determine the net force acting on the crate and then use that to calculate the acceleration. Finally, we can use the acceleration to find the final speed of the crate.

First, let's analyze the forces acting on the crate during each segment of the motion:

1) For the first 11 meters where the floor is frictionless:
Since there is no friction, the only force acting on the crate is the horizontal force applied, which is 450 N.

2) For the next 11 meters where the coefficient of friction is 0.30:
Here, we have two forces acting on the crate:
- The pulling force of 450 N.
- The force of friction opposing the motion.

To find the frictional force, we can use the formula:

Frictional Force = coefficient of friction * Normal force

The normal force is equal to the weight of the crate, which is given by:

Normal force = mass * gravity

where mass = 120 kg and gravity = 9.8 m/s^2.

Now, let's calculate the forces and accelerations:

1) For the first segment (frictionless floor):
Net force = applied force = 450 N

Using Newton's second law of motion, F = m * a, where F is the net force, m is the mass of the crate, and a is the acceleration, we can rearrange the formula to solve for acceleration:

a = F / m
a = 450 N / 120 kg
a = 3.75 m/s^2

2) For the second segment (friction with coefficient of friction 0.30):
Normal force = mass * gravity
Normal force = 120 kg * 9.8 m/s^2
Normal force = 1176 N

Frictional force = coefficient of friction * Normal force
Frictional force = 0.30 * 1176 N
Frictional force = 352.8 N

Net force = applied force - frictional force
Net force = 450 N - 352.8 N
Net force = 97.2 N

Using the formula F = m * a, we can solve for acceleration:

a = F / m
a = 97.2 N / 120 kg
a = 0.81 m/s^2

Now that we have the acceleration, we can find the final speed of the crate using the equation for constant acceleration:

v^2 = u^2 + 2 * a * s

where v is the final speed, u is the initial speed (which is 0 m/s since the crate starts from rest), a is the acceleration, and s is the distance traveled.

For the first segment (frictionless floor):
s = 11 m
v^2 = 0^2 + 2 * 3.75 m/s^2 * 11 m
v^2 = 0 + 2 * 41.25 m^2/s^2
v^2 = 82.5 m^2/s^2
v = √82.5 m/s
v ≈ 9.09 m/s

For the second segment (friction with coefficient of friction 0.30):
s = 11 m
v^2 = 0 + 2 * 0.81 m/s^2 * 11 m
v^2 = 0 + 2 * 8.91 m^2/s^2
v^2 = 17.82 m^2/s^2
v = √17.82 m/s
v ≈ 4.22 m/s

Therefore, the final speed of the crate is approximately 4.22 m/s.