Problem 32 - College 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 coefficients of kinetic and static friction between the box and the surface are 0.200 and \(0.450,\) respectively. (a) What horizontal force must the worker apply to maintain the motion of the box? (b) If the worker stops pushing, what will be the acceleration of the box?

Step-by-Step Solution

Answer

(a) 21.97 N to maintain constant speed. (b) -1.96 m/s² when worker stops pushing.

Step 1: Identify Given Quantities

We are given the mass of the box as \( m = 11.2 \) kg, the coefficient of kinetic friction \( \mu_k = 0.200 \), and the coefficient of static friction \( \mu_s = 0.450 \). The box moves at a constant speed of \( 3.50 \) m/s.

Step 2: Calculate the Normal Force

The normal force \( N \) for a box on a horizontal surface is equal to the weight of the box. This can be calculated as \( N = mg \), where \( g = 9.81 \) m/s² is the acceleration due to gravity. Thus, \( N = 11.2 \times 9.81 = 109.872 \) N.

Step 3: Calculate the Frictional Force

Since the box is moving, we use the coefficient of kinetic friction. The frictional force \( f_k \) is given by \( f_k = \mu_k N \). Substituting the values, \( f_k = 0.200 \times 109.872 = 21.9744 \) N.

Step 4: Determine the Horizontal Force Required

To keep the box moving at a constant speed, the horizontal force applied by the worker must be equal to the frictional force. Thus, the force \( F_{applied} = 21.9744 \) N.

Step 5: Calculate the Acceleration If Pushing Stops

When the worker stops pushing, the only horizontal force acting on the box is the frictional force, which will act in the opposite direction to motion. The box will decelerate due to this force. Use \( f_k = ma \) to find the acceleration \( a \), thus \( a = \frac{f_k}{m} = \frac{21.9744}{11.2} = 1.962\) m/s² in the opposite direction to motion.

Key Concepts

Kinetic Friction

Kinetic friction is the force that opposes the movement of two surfaces sliding past each other. It plays a vital role in everyday activities, such as walking or driving, as it helps control motion. In the exercise, we have a box that is already moving, meaning kinetic friction is at play.

To calculate the kinetic frictional force, you use the formula:

• \[ f_k = \mu_k N \]

Here, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force. In this problem, the coefficient of kinetic friction is given as 0.200, and the normal force, which is the force acting perpendicular to the surfaces in contact, has been calculated as 109.872 N. Thus, the kinetic frictional force is 21.9744 N. This force is crucial because it determines how much force is needed to maintain the box's movement.

Understanding the concept of kinetic friction helps in solving problems related to motion control and energy dissipation in moving objects.

Static Friction

Static friction is the force that prevents two surfaces from sliding past one another when they are at rest or just about to move. It must be overcome to start the motion. In our exercise, knowing the coefficient of static friction helps us see how much force would have been needed to initiate the box's movement.

Before the box starts moving, the static frictional force can be calculated with:

• \[ f_s = \mu_s N \]

In this case, the static friction coefficient \( \mu_s \) is 0.450. Since the normal force remains the same at 109.872 N, the maximum static friction would be 49.4424 N. This is higher than the kinetic friction because it generally takes more force to start moving an object at rest than to keep it moving. Static friction is essential in ensuring objects stay in place until a significant enough force is applied.

Horizontal Force

The horizontal force is the push or pull required to move an object along a straight line on a surface. In the exercise, the worker needs to apply a horizontal force equal to the kinetic frictional force to maintain the constant speed of the moving box.

The calculation involved is simple since the force needed to maintain motion is the same as the frictional force opposing the movement. We found that the applied horizontal force must be 21.9744 N. Therefore, this is the exact force needed to balance out the friction and keep the box sliding at a constant speed of 3.50 m/s.

Horizontal forces play a role not just in keeping objects moving, but also in other activities like sports, vehicles starting from a rest position, or objects being manually pulled or pushed across surfaces.

Acceleration Calculation

When discussing acceleration, we're talking about the change in velocity of an object over time. It's a fundamental concept in motion analysis, allowing us to understand how quickly or slowly objects are increasing or decreasing their speed.

In the provided exercise, once the worker stops pushing, the box experiences deceleration due to the kinetic friction. The only horizontal force acting is the frictional force opposing the movement, and this is used to calculate the box's acceleration as follows:

• \[ a = \frac{f_k}{m} \]

Plugging in the values, the acceleration when the box is no longer pushed is \( 1.962 \) m/s². It's important to note that this acceleration is in the opposite direction of the box's motion, thereby reducing its speed. Calculating acceleration accurately is key in making precise predictions in physics about how objects behave over time.