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When objects slide across a surface, they encounter a resisting force known as kinetic friction. This force acts in the opposite direction to the motion. In our problem, the worker is pushing a 30 kg crate across a floor. The coefficient of kinetic friction, represented as \(\mu_k\), is given as 0.25. This value is essential for calculating how much force one needs to keep the crate moving at a constant velocity. The kinetic friction is calculated using the formula:

• \(F_{f} = \mu_k \times F_n\)

Here, \(F_f\) is the force of friction, and \(F_n\) is the normal force. Since the crate is moving along a level surface, the normal force equates to the crate's weight. Thus, to maintain a constant velocity, the worker must counteract this frictional force by applying an equal and opposite force.

To find out the magnitude of the force the worker needs to apply, we first calculate the normal force. The normal force is approximately the force exerted by the floor to support the weight of the crate. It can be calculated using the formula:

• \(F_n = m \times g\)

Where \(m\) is the mass of the crate (30 kg), and \(g\) is the acceleration due to gravity (approximated to 9.8 m/s虏). This yields \(F_n = 294 \text{ N}\).
The frictional force is then:

• \(F_f = \mu_k \times F_n = 0.25 \times 294 \text{ N} = 73.5 \text{ N}\)

Thus, the worker must apply a force of 73.5 N to keep the crate moving steadily.

Work in physics is defined as the force applied to an object multiplied by the distance over which the force is applied. The formula to calculate work (\(W\)) is:

• \(W = F \times d \times \cos(\theta)\)

Since the worker's force is in the same direction as the movement, \(\theta = 0\), making \(\cos(\theta) = 1\). Thus, the work done by the worker is:
\(W_w = 73.5 \text{ N} \times 4.5 \text{ m} = 330.75 \text{ J}\).
The work done by friction opposes this, given by:
\(W_f = -73.5 \text{ N} \times 4.5 \text{ m} = -330.75 \text{ J}\).
Since friction acts in the opposite direction, its work is negative, cancelling out the work done by the applied force.

The normal force is the perpendicular force exerted by a surface to support the weight of an object resting upon it. In this scenario, the normal force corresponds to the weight of the crate, preventing it from sinking into the floor. Neither the normal force nor gravity does work in this problem since they are both perpendicular to the crate's horizontal motion. Work is associated only with the component of force in the direction of movement, and in this case:

• \(W_n = 0 \text{ J}\)
• \(W_g = 0 \text{ J}\)

Both these components contribute zero to the total work done on the crate, which ensures the problem鈥檚 net total work remains zero.鈥�