Problem 70 - College Physics

Question

How much power must you exert to horizontally drag a \(25.0-\mathrm{kg}\) table \(10.0 \mathrm{~m}\) across a brick floor in \(30.0 \mathrm{~s}\) at constant velocity, assuming the coefficient of kinetic friction between the table and floor is \(0.550 ?\)

Step-by-Step Solution

Answer

You must exert approximately 44.92 watts of power.

Step 1: Identify Given Values

We have:- Mass of the table, \( m = 25.0 \, \mathrm{kg} \)- Distance, \( d = 10.0 \, \mathrm{m} \)- Time, \( t = 30.0 \, \mathrm{s} \)- Coefficient of kinetic friction, \( \mu_k = 0.550 \)Objective: Find the power exerted.

Step 2: Calculate Normal Force

The normal force, \( F_n \), on a horizontal surface is equal to the gravitational force, which is \( F_n = mg \).Calculate it:\[F_n = 25.0 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 245 \, \mathrm{N}\]

Step 3: Calculate Frictional Force

The frictional force, \( F_f \), is given by the formula \( F_f = \mu_k F_n \).Substitute the values:\[F_f = 0.550 \times 245 \, \mathrm{N} = 134.75 \, \mathrm{N}\]

Step 4: Determine the Work Done Against Friction

Work done by friction, \( W \), is calculated as \( W = F_f d \).Calculate it:\[W = 134.75 \, \mathrm{N} \times 10.0 \, \mathrm{m} = 1347.5 \, \mathrm{J}\]

Step 5: Calculate Power Exerted

Power, \( P \), is the work done per unit time, \( P = \frac{W}{t} \).Calculate the power:\[P = \frac{1347.5 \, \mathrm{J}}{30.0 \, \mathrm{s}} \approx 44.92 \, \mathrm{W}\]

Key Concepts

Kinetic Friction

Kinetic friction is a force that opposes the motion of two surfaces sliding past each other. When an object, like a table, is dragged across a floor, kinetic friction arises between the object’s base and the floor surface. This friction depends on:

• The normal force: This is typically the object's weight when on a flat surface, calculated using mass multiplied by gravity.
• The coefficient of kinetic friction: This is a dimensionless number that represents how much frictional force exists between the two surfaces. In our problem, it's given as 0.550.

The formula to find the frictional force, given the above parameters, is:\[ F_f = \mu_k \times F_n \]Where:

• \( \mu_k \): Coefficient of kinetic friction
• \( F_n \): Normal force

Understanding kinetic friction is crucial as it directly affects physical phenomena such as stopping distances in vehicles, efficiency in systems, and is a factor in many engineering calculations.

Work and Energy

Work and energy are closely related physics concepts. Work is done when a force causes an object to move in the direction of the force. It's calculated using:\[ W = F \times d \]Where:

• \( W \): Work done
• \( F \): Force applied
• \( d \): Distance over which the force is applied

In terms of energy, doing work on an object transfers energy to it or changes its form. For example, when you perform work by dragging an object, such as a table, the applied energy is overcoming friction and moving the object forward.
In our exercise, we specifically focus on the work done against the force of kinetic friction. As you pull the table across the floor, your exertion results in the table moving a certain distance, using up energy. Calculating this work helps us understand how much effort is needed to maintain motion over a specified distance.

Constant Velocity Motion

Constant velocity means that an object moves at a uniform speed in a straight line. In this condition, the speed and direction of the object do not change.
To maintain constant velocity while dragging the table, the force you apply must exactly balance the force of kinetic friction. This means the net force acting on the table is zero:

• No acceleration: As acceleration involves a change in velocity, it’s not present when velocity is constant.
• Equilibrium of forces: In this state, the pulling force equals the frictional force.

Solving for power in these conditions shows that while energy is expended at a constant rate, it translates to maintaining speed rather than increasing it. Power, which is the rate of doing work, in this context, helps us quantify how much effort needs to be sustainable over time to keep that constant velocity.