Question

How much work do movers do (horizontally) in pushing a \(150-\mathrm{kg}\) crate \(12.3 \mathrm{~m}\) across a floor at constant speed if the coefficient of friction is \(0.70 ?\) a) 1300 J b) 1845 J c) \(1.3 \cdot 10^{4}\). d) \(1.8 \cdot 10^{4}\) J e) \(130 \mathrm{~J}\)

Short answer

a) \(7.4 \cdot 10^{3} \mathrm{J}\) b) \(1.0 \cdot 10^{4} \mathrm{J}\) c) \(1.3 \cdot 10^{4} \mathrm{J}\) d) \(1.5 \cdot 10^{4} \mathrm{J}\) Answer: c) \(1.3 \cdot 10^{4} \mathrm{J}\)

Step-by-step solution

Calculate the weight of the crate

To calculate the weight of the crate, we need to multiply its mass by the acceleration due to gravity (\(g \approx 9.81 \mathrm{m/s^2}\)). Therefore, Weight = Mass × Gravity Weight = \(150 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 1471.5 \mathrm{N}\)

Calculate the force of friction

To calculate the force of friction, we need to multiply the weight of the crate by the coefficient of friction: Force of friction = Coefficient of friction × Weight Force of friction = \(0.70 \times 1471.5 \mathrm{N} = 1029.05 \mathrm{N}\)

Calculate the work done by the movers

To calculate the work done by the movers, we need to multiply the force of friction by the distance over which the crate is moved: Work done = Force of friction × Distance Work done = \(1029.05 \mathrm{N} \times 12.3 \mathrm{m} = 12657.315 \mathrm{J}\)

Choose the correct answer from the given options

We can see that the calculated work done by the movers is approximately \(1.3 \times 10^4 \mathrm{J}\). Therefore, the correct answer is: c) \(1.3 \cdot 10^{4} \mathrm{J}\).

Key concepts

Coefficient of Friction

The coefficient of friction is a straightforward but crucial concept in physics. It's a dimensionless number that represents the amount of friction between two surfaces. When two surfaces are in contact, the coefficient of friction determines the amount of resistance one surface or object encounters when moving over another.

It's calculated by dividing the force of friction between two surfaces by the normal force pressing them together. A higher coefficient means more friction. For instance, a smooth, icy surface may have a low coefficient of friction, allowing objects to move freely across it. On the other hand, rubber on concrete would typically have a higher coefficient, indicating more resistance and less sliding. In the example of the moving crate, the fairly high coefficient of 0.70 indicates significant resistance between the crate and the floor. This factor directly influences the work done by the movers in sliding the crate, as more force would be required to overcome this resistance.

Force Calculation

When calculating force in the context of work done, it's essential to apply Newton's second law of motion, which tells us that force is the product of mass and acceleration: \( F = m \cdot a \).

However, when applying this to work done against friction, as in the case of the movers pushing the crate, the force we're interested in is the force of friction. This force is determined not by the acceleration of the crate, but by its weight (mass times the acceleration due to gravity) and the coefficient of friction of the surface it's moving across. By multiplying these two values—in our case, the weight of the crate and the coefficient of friction—we can calculate the force of friction that must be overcome to move the crate.

This calculated force is vital for determining the work done as it represents the continuous force the movers need to apply over the distance to move the crate at a constant speed.

Work-Energy Principle

The work-energy principle is a fundamental concept in physics that links the work done on an object to its energy change. In simple terms, work is the energy transferred to or from an object via the application of a force along a displacement. The formula to calculate work is \( W = F \cdot d \cos(\theta) \), where \( W \) is work done, \( F \) is the magnitude of the force applied, \( d \) is the displacement of the object, and \( \theta \) is the angle between the force vector and the direction of motion.

For movers pushing a crate horizontally with a force that is parallel to the displacement, \( \theta \) is 0 degrees, and the cosine term becomes 1, simplifying the work calculation to the product of the force and the distance (\cos(0) = 1). Thus, the work done by the movers equals the force of friction times the distance the crate is pushed across the floor. In the textbook problem, the movers do a significant amount of work in moving the heavy crate, which is clearly illustrated by the substantial value calculated, given the friction and the distance involved.