A factory worker pushes a 30.0 kg crate a distance of 4.5 m along a level floor at constant velocityby pushing horizontally on it. The coefficient of kine

It seems like the question is cut off. To provide a detailed solution, I assume you would like to find the force exerted by the worker and perhaps the frictional force acting on the crate, given that it moves at a constant velocity.

1. Break Down the Problem

1. Determine the mass of the crate, $m = 30.0 \, \text{kg}$.
2. The distance pushed is $d = 4.5 \, \text{m}$.
3. Since the crate is moving at a constant velocity, the net force acting on the crate is zero.
4. We can use the coefficient of kinetic friction, which seems to be missing. We'll denote it as $\mu_k$ for now and will assume some value if needed.

2. Relevant Concepts

1. The force of friction, $f_k$, can be calculated using: $f_k = \mu_k \cdot N$ where $N$ is the normal force.

2. The normal force $N$ is equal to the weight of the crate when on a level surface: $N = m \cdot g$ where $g = 9.81 \, \text{m/s}^2$.

3. Analysis and Detail

1. Calculate the normal force: $N = 30.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 294.3 \, \text{N}$

2. The force of friction will then be $f_k = \mu_k \cdot 294.3 \, \text{N}$

3. Because the crate moves at constant velocity, the horizontal force exerted by the worker, $F$, must balance the frictional force: $F = f_k = \mu_k \cdot 294.3 \, \text{N}$

4. Verify and Summarize

You need the value of $\mu_k$ to determine the actual force exerted by the worker. If you provide this value, we can proceed with the calculation. Otherwise, we can summarize that:

• The force of friction and the force exerted by the worker are equal when moving at constant velocity.
• The normal force acting on the crate is $294.3 \, \text{N}$.

Final Answer

Without a specific value for the coefficient of kinetic friction $\mu_k$, the force exerted by the worker can be expressed as: $F = \mu_k \cdot 294.3 \, \text{N}$ Please provide the coefficient of kinetic friction to complete the calculation.