Problem 70 - University Physics with Modern Physics

Question

A small block with mass 0.0400 kg slides in a vertical circle of radius \(R =\) 0.500 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point \(A\), the normal force exerted on the block by the track has magnitude 3.95 N. In this same revolution, when the block reaches the top of its path, point \(B\), the normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point \(A\) to point \(B\)?

Step-by-Step Solution

Answer

The work done by friction is -1.56 J.

Step 1: Understand the forces involved

At the bottom of the circular path (point A), two forces act on the block: the gravitational force (\( F_g = mg \)) and the normal force exerted by the track. At the top (point B), the forces are the same, but their direction relative to the circular path changes.

Step 2: Calculate the gravitational force

The gravitational force is calculated using \( F_g = mg \), where \( m = 0.0400 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Thus, \( F_g = 0.0400 \, \times 9.8 = 0.392 \, \text{N} \).

Step 3: Determine velocities at points A and B

At point A (bottom), the centripetal force \( F_c = F_N - F_g = 3.95 \, \text{N} - 0.392 \, \text{N} = 3.558 \, \text{N} \). The centripetal force is also given by \( F_c = \frac{mv^2}{R} \), so \( v_A^2 = \frac{3.558 \, \text{N} \times 0.5 \, \text{m}}{0.0400 \, \text{kg}} = 44.475 \, \text{m}^2/\text{s}^2 \), which gives \( v_A = 6.67 \, \text{m/s} \). At point B (top), \( F_c = F_N + F_g = 0.680 \, \text{N} + 0.392 \, \text{N} = 1.072 \, \text{N} \), so \( v_B^2 = \frac{1.072 \, \text{N} \times 0.5 \, \text{m}}{0.0400 \, \text{kg}} = 13.4 \, \text{m}^2/\text{s}^2 \), giving \( v_B = 3.66 \, \text{m/s} \).

Step 4: Calculate work done by friction

The work-energy principle states \( \Delta K = W_{friction} + \Delta U \). The change in kinetic energy (\( \Delta K \)) is \( \frac{1}{2}m(v_B^2 - v_A^2) \). \( \Delta U = mg(2R) \) because the block gains height equal to the diameter of the circle. Substituting the values, \( \Delta U = 0.0400 \, \times 9.8 \, \times 1.0 = 0.392 \, \text{J} \). Finally, solve for work done by friction: \( W_{friction} = \Delta K - \Delta U = \frac{1}{2} \times 0.0400 \, \times (13.4 - 44.475) - 0.392 = -1.556 \, \text{J}. \) The negative sign indicates that work is done against the motion.

Key Concepts

Circular Motion

Circular motion refers to the motion of an object along the circumference of a circle. When an object moves in a circle, like our small block inside a vertical circular track, its direction constantly changes. This continuous change in direction implies that the object is accelerating, even if its speed remains constant.
In circular motion, several forces come into play. The gravitational force always acts downward along the circle's path. When examining the forces at different points, like at the top (Point B) and bottom (Point A) positions of the track, differences in forces become apparent due to their directions.
The forces not only affect the speed but also determine the kind of motion the object will have. These differences are key to analyzing what happens during its motion from bottom to top and back, including how much work is done by forces like friction.

Centripetal Force

Centripetal force is the force that keeps an object moving in a circular path. In simple terms, it acts towards the center of the circle and keeps our block glued to the track. Without this force, the block would just move off in a straight line, thanks to inertia.
This force can be calculated using the formula \( F_c = \frac{mv^2}{R} \). Here, \( m \) is the mass of the object, \( v \) is its velocity, and \( R \) is the radius of the circle. In our example, at point A, the centripetal force results from the difference between the normal force and gravitational force, while at point B, it is the sum of these forces.
Understanding how different forces contribute to the centripetal force helps predict changes in the block's speed at varying points in its journey around the circle.

Work-Energy Principle

The work-energy principle is a fundamental concept that explains the relationship between work and energy change. It tells us that the work done by all forces acting on an object results in a change in its kinetic energy. Mathematically, this is expressed as \( \Delta K = W_{friction} + \Delta U \).
In our exercise, this principle helps calculate the work done by friction as the block moves from point A to point B. The change in kinetic energy (\( \Delta K \)) is calculated using the initial and final speeds at these points. Moreover, the potential energy change (\( \Delta U \)) results from the block moving against gravity as it rises to the top of the circle.
By rearranging the formula, the work done by friction can be found. A negative value indicates energy is removed from the system by friction, opposing the block's motion.