Understanding the Frictional Force Problem

This question asks us to find the magnitude of the frictional force acting on a block moving on a rough horizontal surface. We are given the block's mass, initial speed, and the distance it travels before coming to rest. The frictional force is the force that opposes the motion of the block, causing it to slow down and stop.

Identifying Given Information

Here's what we know from the problem statement:

• Mass of the block (\(m\)) = 2 kg
• Initial speed of the block (\(u\)) = 3 m/s
• Final speed of the block (\(v\)) = 0 m/s (since it comes to rest)
• Distance traveled (\(s\)) = 3 m
• The surface is rough, indicating the presence of frictional force.

We need to find the magnitude of the frictional force (\(f\)).

Key Physics Concepts for Solving Frictional Force

We can solve this problem using either the Work-Energy Theorem or by combining Kinematics and Newton's Second Law. Let's use the Work-Energy Theorem first, as it directly relates the work done by friction to the change in kinetic energy.

• Work-Energy Theorem: The net work done on an object is equal to the change in its kinetic energy. Work_net = \(\Delta KE\).
• Work Done by a Force: Work (\(W\)) = Force (\(F\)) \(\times\) Distance (\(s\)) \(\times \cos(\theta)\), where \(\theta\) is the angle between the force and the displacement.
• Kinetic Energy: \(KE = \frac{1}{2}mv^2\).

Applying the Work-Energy Theorem to Find Frictional Force

The only horizontal force doing work on the block is the frictional force. Gravity and the normal force are vertical and do no work as the displacement is horizontal (\(\theta = 90^\circ\), \(\cos 90^\circ = 0\)).

The frictional force acts opposite to the direction of motion. So, the angle between the frictional force and the displacement is \(180^\circ\), and \(\cos 180^\circ = -1\).

The work done by the frictional force is \(W_{friction} = f \times s \times \cos(180^\circ) = f \times s \times (-1) = -fs\).

The net work done on the block is equal to the work done by friction: \(W_{net} = -fs\).

The change in kinetic energy is \(\Delta KE = KE_{final} - KE_{initial}\).

• Initial Kinetic Energy: \(KE_{initial} = \frac{1}{2}mu^2 = \frac{1}{2}(2 \text{ kg})(3 \text{ m/s})^2 = \frac{1}{2}(2)(9) = 9 \text{ J}\).
• Final Kinetic Energy: \(KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2}(2 \text{ kg})(0 \text{ m/s})^2 = 0 \text{ J}\).

Change in Kinetic Energy: \(\Delta KE = 0 \text{ J} - 9 \text{ J} = -9 \text{ J}\).

According to the Work-Energy Theorem:

\(W_{net} = \Delta KE\)

\(-fs = -9 \text{ J}\)

Substitute the distance traveled (\(s = 3\) m):

\(-f(3 \text{ m}) = -9 \text{ J}\)

\(-3f = -9\)

\(f = \frac{-9}{-3}\)

\(f = 3 \text{ N}\)

The magnitude of the frictional force is 3 N.

Alternative Approach: Kinematics and Newton's Second Law

We can first find the acceleration using kinematics and then use Newton's second law.

Using the kinematic equation \(v^2 = u^2 + 2as\):

• \(v = 0\) m/s
• \(u = 3\) m/s
• \(s = 3\) m

\(0^2 = (3 \text{ m/s})^2 + 2a(3 \text{ m})\)

\(0 = 9 + 6a\)

\(6a = -9\)

\(a = \frac{-9}{6} = -1.5 \text{ m/s}^2\)

The acceleration is -1.5 m/s\(^2\), meaning the block is decelerating at a rate of 1.5 m/s\(^2\).

Now, apply Newton's Second Law (\(F_{net} = ma\)). The only horizontal force is the frictional force \(f\), acting in the direction opposite to the initial motion. If we take the initial direction of motion as positive, the force is \(-f\).

\(F_{net} = ma\)

\(-f = (2 \text{ kg})(-1.5 \text{ m/s}^2)\)

\(-f = -3 \text{ N}\)

\(f = 3 \text{ N}\)

Both methods yield the same magnitude for the frictional force: 3 N.

Summary of Frictional Force Calculation

Based on the given values and applying physics principles, the magnitude of the frictional force causing the 2 kg block moving at 3 m/s to stop over 3 m is 3 N.

Revision Table: Frictional Force Problem

Additional Information: Types of Frictional Force

Friction is a force that opposes relative motion or the tendency of motion between surfaces in contact. There are different types of friction:

• Static Friction: This is the friction that prevents an object from starting to move when a force is applied. Its magnitude varies from zero up to a maximum value (\(f_s \le \mu_s N\)), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force.
• Kinetic Friction (or Sliding Friction): This is the friction that opposes the motion of an object that is sliding over a surface. Its magnitude is generally constant (\(f_k = \mu_k N\)), where \(\mu_k\) is the coefficient of kinetic friction and \(N\) is the normal force. In this problem, since the block is moving, we are dealing with kinetic friction.
• Rolling Friction: This is the friction that opposes the motion of a rolling object. It is generally much smaller than kinetic friction.
• Fluid Friction (or Drag): This is the friction force exerted by a fluid (like air or water) on an object moving through it.

In this specific problem, the frictional force calculated is the magnitude of the kinetic friction acting on the block as it slides to a stop.