A baggage handler throws a 16.0 kg suitcase horizontally along the floor of an airplane luggage compartment with an initial speed of 1.10 m/s . The suitcase slides 2.10 m before stopping.

To solve this problem, we can use the principles of Newton's second law and the work-energy theorem. Let's break it down step by step.

Step 1: Identify the known quantities
- Mass of the suitcase (m): 16.0 kg
- Initial velocity (v₀): 1.10 m/s
- Distance traveled (d): 2.10 m

Step 2: Calculate the final velocity (v)
Since the suitcase eventually stops, its final velocity (v) will be zero.
v = 0 m/s

Step 3: Calculate the acceleration (a)
We can use the equation of motion to find the acceleration using the initial velocity (v₀), final velocity (v), and distance traveled (d):
v² = v₀² + 2ad

Rearranging the equation, we can solve for acceleration:
a = (v² - v₀²) / (2d)
= (0² - 1.10²) / (2 * 2.10)

Step 4: Calculate the acceleration (a)
a = (-1.21) / 4.20
= -0.29 m/s²

Step 5: Calculate the force (F) applied by the baggage handler
Using Newton's second law (F = ma), we can find the force applied:
F = m * a
= 16.0 kg * (-0.29 m/s²)

Step 6: Calculate the force (F)
F = -4.64 N

So, the baggage handler applies a force of -4.64 N to stop the suitcase.

To solve this problem, we can use the principles of physics, specifically the laws of motion and the concept of work.

First, let's calculate the initial kinetic energy (K_i) of the suitcase. Kinetic energy is given by the formula:

K_i = 1/2 * m * v_i^2

where:
m is the mass of the suitcase (16.0 kg)
v_i is the initial velocity of the suitcase (1.10 m/s)

Substituting the given values into the formula:

K_i = 1/2 * 16.0 kg * (1.10 m/s)^2
K_i = 1/2 * 16.0 kg * 1.21 m^2/s^2
K_i = 9.76 J

Now, let's calculate the work (W) done by the friction force to stop the suitcase. The work done by friction is equal to the change in kinetic energy. Since the suitcase comes to a stop, the final kinetic energy (K_f) is zero. Therefore:

W = K_f - K_i

Substituting the given values:

W = 0 J - 9.76 J
W = -9.76 J

The negative sign indicates that work has been done on the suitcase by the friction force.

Since the friction force over a sliding distance is equal to the work done divided by the distance, we can calculate the friction force (F) as follows:

F = W / d

where:
W is the work done (-9.76 J)
d is the distance the suitcase slides (2.10 m)

Substituting the given values:

F = -9.76 J / 2.10 m
F = -4.65 N

The negative sign indicates that the friction force acts in the opposite direction to the motion of the suitcase.

Therefore, the baggage handler applied a force of approximately 4.65 N in the opposite direction to the motion of the suitcase to bring it to a stop.

Thanks. Is there a question here? Such as what is the coefficent of friction?

Vf^2=Vi^2+ 2ad where a=mu*mg/m=mu*g

0=1.1^2+2mu*g *"(- 2.10_
solve for mu