A 120-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 400 N. For the first 17 m the floor is frictionless, and for the next 17 m the coefficient of friction is 0.34. What is the final speed of the crate?

Kinetic energy gained during the frictionless interval is 400 N * 17 m = 6800 J. When there is friction, the net force pulling the weight is 400 N - M*g*mu = 400 - 399.8 N = 0.2 N. Additional kinetic energy gain is limited to 0.2N*17m = 3.4 J

The rest of the work done ends up as heat.

Final kientic energy is 6803 J. Use that for the final velocity

how would I show that in m/s?

you use 1/2*m*v^2

the answer is = 1

To find the final speed of the crate, we need to break down the problem into two parts: the frictionless section and the section with friction.

1. Frictionless section:
In this part, the floor is frictionless, so there is no force opposing the motion of the crate. Therefore, the only force acting on the crate is the applied force pulling it with a constant horizontal force of 400 N. We can use Newton's second law of motion to find the crate's acceleration in this section.

The formula for Newton's second law is:
F = m * a

Where:
F is the net force on the object
m is the mass of the object
a is the acceleration of the object

In this case, the only force acting on the crate is the applied force, so we have:
400 N = 120 kg * a

Solving for the acceleration:
a = 400 N / 120 kg ≈ 3.33 m/s^2

Now, we can find the final velocity in this section using the following kinematic equation:
v^2 = v0^2 + 2 * a * d

Where:
v is the final velocity
v0 is the initial velocity (which is zero in this case, since the crate starts from rest)
a is the acceleration (3.33 m/s^2)
d is the distance traveled (17 m)

Plugging in the values:
v^2 = 0 + 2 * 3.33 m/s^2 * 17 m
v^2 ≈ 113.22

Taking the square root of both sides:
v ≈ √113.22
v ≈ 10.65 m/s

So, the final velocity of the crate in the frictionless section is approximately 10.65 m/s.

2. Section with friction:
In this part, the coefficient of friction is 0.34. First, let's calculate the frictional force acting on the crate using the equation:

frictional force = coefficient of friction * normal force

The normal force can be calculated as:
normal force = mass * gravity

Where:
mass is the mass of the crate (120 kg)
gravity is the acceleration due to gravity (approximately 9.8 m/s^2)

normal force = 120 kg * 9.8 m/s^2 ≈ 1176 N

frictional force = 0.34 * 1176 N ≈ 399.84 N

The frictional force is opposing the motion, so it acts in the opposite direction to the applied force. Hence, the net force acting on the crate in this section can be calculated as:
net force = applied force - frictional force
net force = 400 N - 399.84 N ≈ 0.16 N

Since the net force is very close to zero, the acceleration in this section is negligible. Therefore, the velocity remains constant.

So, the final velocity of the crate in the section with friction is the same as its velocity in the frictionless section, which is approximately 10.65 m/s.

Therefore, the final speed of the crate is approximately 10.65 m/s.