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 c) \(1.3 \cdot 10^{4}\) ] e) 130 ] b) 1845 J d) \(1.8 \cdot 10^{4}\) ]

Short answer

The distance travelled is 12.3 meters, and the coefficient of friction between the floor and the crate is 0.70. Answer: (c) \(1.3 \cdot 10^{4} \mathrm{J}\)

Step-by-step solution

Determine the force of friction

Since the crate is moving at a constant speed, it means the force exerted by the movers is equal to the force of friction. To calculate the force of friction, we can use the following equation: _force of friction = coefficient of friction × normal force_ The normal force in this case is equal to the gravitational force acting on the crate, which can be calculated as follows: _normal force = mass × acceleration due to gravity_ Where mass is given as \(150 \mathrm{kg}\), and acceleration due to gravity can be taken as \(9.81 \mathrm{m/s^2}\). Let's calculate the normal force and then the force of friction.

Calculate the normal force

Using the formula for normal force, we get: _normal force = 150 \mathrm{kg} × 9.81 \mathrm{m/s^2} = 1471.5 \mathrm{N}_ Now, we have the normal force needed to calculate the force of friction.

Calculate the force of friction

Now, we can calculate the force of friction using the given coefficient of friction and the calculated normal force. _force of friction = 0.70 × 1471.5 \mathrm{N} = 1029.05 \mathrm{N}_ The force of friction is equal to the force exerted by the movers.

Calculate the work done

To calculate the work done by the movers, we can use the following equation: _work done = force × distance × cos(theta)_ Since the movers are pushing the crate horizontally, the angle (theta) would be 0 degrees, and cos(0) = 1. We have calculated the force exerted by the movers and the distance travelled is given as \(12.3 \mathrm{m}\). Now, let's plug in these values to calculate the work done.

Calculate the work done by the movers

Using the formula for work done, we get: _work done = 1029.05 \mathrm{N} × 12.3 \mathrm{m} × 1 = 12656.815 \mathrm{J}_ The work done by the movers is approximately \(1.3 × 10^4 \mathrm{J}\). The correct answer is (c) \(1.3 \cdot 10^{4} \mathrm{J}\).

Key concepts

Friction

Friction is a force that opposes the motion of objects as they slide past each other. It plays a crucial role in everyday activities, from walking to driving. The force of friction depends on two factors: the nature of the surfaces in contact and the normal force pressing them together.
The coefficient of friction, denoted as \( \mu \), is a dimensionless number that represents the degree to which a surface resists sliding. In our example, the coefficient of friction is given as 0.70. This means that the frictional force is 70% of the normal force.
The frictional force can be calculated using the equation:

• \( \text{frictional force} = \mu \times \text{normal force} \)

This helps us understand how much effort is needed to move an object against frictional resistance.

Normal Force

Normal force is an essential concept in physics that represents the perpendicular force exerted by a surface upon an object lying on it. It acts at a right angle to the surface.
For an object resting on a horizontal surface, the normal force is equal to the weight of the object. Hence, it can be calculated using the formula:

• \( \text{normal force} = \text{mass} \times \text{acceleration due to gravity} \)

In our scenario, the mass of the crate is 150 kg, and the acceleration due to gravity is approximately 9.81 m/s². Plugging these values into the formula gives us a normal force of 1471.5 N.
It is important to note that the normal force can vary if the object is on an inclined plane, but it remains equal to the weight of the object on a flat horizontal surface.

Constant Speed

When an object is said to be moving at a constant speed, it means that its speed does not change over time. In physics, this is associated with a balance of forces.
In the exercise we have, the crate is pushed at a constant speed. This implies that the force exerted by the movers (pushing force) is equal to the opposing frictional force.
Achieving constant speed tells us that there is equilibrium, where the net force acting on the object is zero. Therefore, no acceleration is happening, even though work is being done. This plays a crucial role in determining the work done as it simplifies calculations since only the force needed to counteract friction is considered.

Work Equation

Work is a measure of energy transfer that occurs when an object is moved over a distance by an external force. The work equation is given by:

• \( \text{work} = \text{force} \times \text{distance} \times \cos(\theta) \)

where \( \theta \) is the angle between the direction of the force and the direction of movement.
In this example, since the force applied is in the same direction as the movement of the crate, the angle \( \theta \) is 0 degrees. The cosine of 0 degrees is 1, making the equation simpler:

• \( \text{work} = \text{force} \times \text{distance} \)

Plugging the values for force (1029.05 N) and distance (12.3 m) into the equation gives us the total work done by the movers, which is calculated to be approximately 12656.815 J, or \( 1.3 \times 10^4 \) J.