Problem 27 - University Physics with Modern Physics

Question

A stockroom worker pushes a box with mass 11.2 \(\mathrm{kg}\) on a horizontal surface with a constant speed of 3.50 \(\mathrm{m} / \mathrm{s}\) . The coefficient of kinetic friction between the box and the surface is 0.20 (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

Step-by-Step Solution

Answer

(a) The worker must apply a force of 22 N. (b) The box slides 3.12 meters before stopping.

Step 1: Understanding the Problem

We know that the box is moving at a constant speed, which implies that the net force acting on it is zero. The worker applies a force to balance out the frictional force to maintain this motion. We need to first calculate the force needed to balance the kinetic friction.

Step 2: Calculating the Force of Kinetic Friction

The force of kinetic friction can be calculated using the formula: \[ f_k = \mu_k \cdot N \]where \( \mu_k = 0.20 \) (the coefficient of kinetic friction) and \( N \) is the normal force, which is equal to the gravitational force in this case because the surface is horizontal. Thus, \( N = m \cdot g = 11.2 \text{ kg} \times 9.81 \text{ m/s}^2 \).

Step 3: Compute the Normal Force

The normal force \( N \) is calculated as follows:\[ N = 11.2 \text{ kg} \times 9.81 \text{ m/s}^2 = 109.872 \text{ N} \]

Step 4: Calculate the Kinetic Frictional Force

Substitute the normal force into the friction equation:\[ f_k = 0.20 \times 109.872 \text{ N} = 21.9744 \text{ N} \]This is the frictional force the worker must balance.

Step 5: Calculating the Horizontal Force Required (Part a)

To move the box at constant speed, the worker needs to apply a horizontal force equal to the kinetic frictional force:\[ F = f_k = 21.9744 \text{ N} \]Thus, the horizontal force required is approximately 22 N.

Step 6: Understanding Removal of Force

Now, if the force is removed, the box will start decelerating due to the frictional force. We need to calculate how far it moves until it stops.

Step 7: Calculating the Deceleration Due to Friction

When the force is removed, only the friction acts on the box. The acceleration can be found using:\[ a = \frac{-f_k}{m} = \frac{-21.9744 \text{ N}}{11.2 \text{ kg}} \]

Step 8: Compute the Acceleration

Substitute the values:\[ a = \frac{-21.9744}{11.2} = -1.962 \text{ m/s}^2 \]This is the deceleration of the box.

Step 9: Use Kinematic Equation to Find Distance (Part b)

We apply the kinematic equation:\[ v^2 = u^2 + 2as \]With \( v = 0 \) (final velocity), \( u = 3.5 \text{ m/s} \) (initial velocity), and \( a = -1.962 \text{ m/s}^2 \). We solve for \( s \), the distance.

Step 10: Calculate the Distance

Plugging in the values into the kinematic equation:\[ 0 = (3.5)^2 + 2(-1.962)s \]\[ s = \frac{(3.5)^2}{2 \times 1.962} \approx 3.12 \text{ m} \]Therefore, the box slides approximately 3.12 meters before coming to rest.

Key Concepts

Normal Force

The normal force is an essential concept when discussing motion on surfaces. It is the force exerted by a surface to support the weight of an object resting on it. In this scenario, the stockroom worker deals with a box on a horizontal surface. As the box rests on this surface, gravity pulls it downward. The surface responds by exerting an upward normal force.

To calculate the normal force, we use the formula:

• \[ N = m \cdot g \]

where \( m \) is the mass of the box (11.2 kg) and \( g \) is the acceleration due to gravity (approximately 9.81 \( \text{m/s}^2 \)). In this case, the normal force equals the weight of the box because there are no other vertical forces acting on it. Thus, the normal force is 109.872 N.

This normal force plays a crucial role in determining the force of kinetic friction. Since kinetic friction depends on both the coefficient of friction and the normal force, understanding this concept helps in solving related motion problems effectively.

Horizontal Force

In physics, the horizontal force is crucial in understanding how objects move across surfaces. When an object, like the box in the exercise, moves at a constant speed, it implies that the applied horizontal force balances the frictional force. This means that the net force acting on the box is zero.

The worker must apply a force equal to the force of kinetic friction to maintain constant motion. Kinetic friction can be calculated using the formula:

• \[ f_k = \mu_k \cdot N \]

where \( \mu_k \) is the coefficient of friction (0.20), and \( N \) is the normal force (109.872 N).

In this exercise, the horizontal force required is found to be approximately 22 N. This understanding assists in realizing how different forces interact to result in motion and equilibrium on a plane.

Deceleration

When an object in motion experiences a decrease in speed, it is referred to as deceleration, a form of negative acceleration. In our stockroom worker problem, when the applying force is removed, the box will begin to slow down due to the kinetic friction force opposing its motion.

Deceleration is calculated using Newton's second law of motion, where the force of friction is acting against the mass of the object to decelerate it:

• \[ a = \frac{-f_k}{m} \]

With the kinetic force \( f_k \) as 21.9744 N and the box's mass as 11.2 kg, the box experiences a deceleration of \(-1.962 \, \text{m/s}^{2}\).

This concept is vital for calculating the stopping distance or analyzing how quickly a moving object comes to rest under certain forces.

Kinematic Equations

Kinematic equations are essential tools in physics for predicting the motion of objects. They relate the quantities of velocity, acceleration, displacement, and time. When solving the second part of the exercise, these equations become particularly handy to find how far the box slides after the force is removed.

The equation used here is:

• \[ v^2 = u^2 + 2as \]

where \( v \) is the final velocity (0 m/s, since the box stops), \( u \) is the initial velocity (3.5 m/s), \( a \) is the deceleration (-1.962 m/s²), and \( s \) is the distance traveled.

Solving the equation gives a distance \( s \) of approximately 3.12 meters before the box comes to rest. Understanding how to manipulate these equations allows us to unravel more complex problems related to motion.