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? Express your answer using two significant figures and m/s.

Final kinetic energy = (work done) - (work lost due to friction)

(1/2)M*Vf^2 = 400N*34 m - M*g*Uk*(17 m) = 13,600 J - 6797 J
Solve for Vf.

There is almost no acceleration during the final 17 m. The pulling force nearly equals the friction force. This is a coincidence; they set it up that way

To find the final speed of the crate, we need to break down the problem into two parts: the frictionless section (17 m) and the section with friction (17 m).

First, let's calculate the work done during the frictionless section using the formula:

Work = Force × Distance

During the frictionless section, the crate experiences no friction, so the only force acting on it is the applied force of 400 N. Thus, the work done is:

Work = 400 N × 17 m = 6800 J

Since work is equal to the change in kinetic energy (KE), we can find the KE at the end of the frictionless section:

KE_1 = Work = 6800 J

Next, let's calculate the work done during the section with friction. The work done against friction is given by the equation:

Work = Force of Friction × Distance

The force of friction can be calculated using the equation:

Force of Friction = Coefficient of Friction × Normal Force

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

Weight = mass × gravity

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

Weight = 120 kg × 9.8 m/s^2 = 1176 N

Now, we can calculate the force of friction:

Force of Friction = 0.34 × 1176 N = 399.84 N (approx. 400 N)

The work done against friction is:

Work = 400 N × 17 m = 6800 J

Again, this work done is equal to the change in kinetic energy (KE) during the section with friction:

KE_2 = Work = 6800 J

Now, let's find the total kinetic energy (KE_total) at the end of the second section (the end of the 34 m distance):

KE_total = KE_1 + KE_2 = 6800 J + 6800 J = 13600 J

Finally, we can find the final speed (v) of the crate using the kinetic energy formula:

KE_total = (1/2) × mass × v^2

Substituting the given values:

13600 J = (1/2) × 120 kg × v^2

Simplifying:

v^2 = (2 × 13600 J) / 120 kg = 226.67 m^2/s^2

Taking the square root:

v ≈ √(226.67 m^2/s^2) ≈ 15.04 m/s

Therefore, the final speed of the crate is approximately 15.04 m/s.

To find the final speed of the crate, we need to consider the two parts of its motion: the frictionless section and the section with friction.

First, let's calculate the work done on the crate during the first 17 m on the frictionless floor. Since there is no friction, the only force acting on the crate is the applied force of 400 N. The work done is given by the equation:

Work = Force x Distance x cos(θ)

Since the applied force is horizontal and the angle θ between the force and displacement is 0 degrees, cos(0) = 1. Therefore, the work done can be calculated as:

Work = 400 N x 17 m x 1 = 6800 J

Next, let's calculate the work done against friction during the next 17 m, where the coefficient of friction is 0.34. The frictional force can be calculated using the equation:

Frictional Force = coefficient of friction x Normal force

The normal force can be calculated as the weight of the crate:

Normal force = mass x gravitational acceleration = 120 kg x 9.8 m/s^2 = 1176 N

Therefore, the frictional force is:

Frictional Force = 0.34 x 1176 N = 399.84 N

The work done against friction can be calculated as:

Work = Force x Distance x cos(θ)

In this case, the force is the frictional force and the angle θ between the force and displacement is 180 degrees, cos(180) = -1. Therefore, the work done is:

Work = -399.84 N x 17 m x -1 = 6797.28 J

Now, let's calculate the net work done on the crate. The net work is the sum of the work done on the crate during the frictionless section and the work done against friction:

Net Work = Work (frictionless) + Work (friction) = 6800 J + 6797.28 J = 13597.28 J

The net work done on an object is equal to the change in its kinetic energy. Therefore, we can equate the net work to the change in kinetic energy:

Net Work = Change in Kinetic Energy

The initial kinetic energy of the crate is zero as it starts from rest. Therefore:

Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy

Since the crate ends up with a final speed, the final kinetic energy can be calculated using the equation:

Final Kinetic Energy = 0.5 x mass x (final speed)^2

Plugging in the values, we have:

13597.28 J = 0.5 x 120 kg x (final speed)^2

Solving for the final speed:

(final speed)^2 = (13597.28 J) / (0.5 x 120 kg)
(final speed)^2 = 113.31 m^2/s^2
final speed = √113.31 m^2/s^2
final speed ≈ 10.65 m/s

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