/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 cp A small block with mass 0.040... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The work done by friction is equal to the change in total mechanical energy, including potential energy.

Step by step solution

01

Identify Known Variables

First, we record the known values: mass of the block, \( m = 0.0400 \; \text{kg} \), radius of the circle, \( R = 0.500 \; \text{m} \), gravitational acceleration \( g = 9.81 \; \text{m/s}^2 \), normal force at point \( A \), \( F_{N_A} = 3.95 \; \text{N} \), and normal force at point \( B \), \( F_{N_B} = 0.680 \; \text{N} \).
02

Calculate Kinetic Energy at Point A

At point \( A \), the forces involved are gravitational force and normal force. The net force provides the centripetal acceleration needed for circular motion (\( v_A^2 = R \cdot a_c \)). We use:\[ F_{N_A} = m \cdot g + \frac{m \cdot v_A^2}{R} \]Solving for velocity \( v_A \):\[ 3.95 = 0.0400 \times 9.81 + \frac{0.0400 \times v_A^2}{0.500} \]Solve this equation to find \( v_A \).
03

Calculate Velocity at Point A

Rearranging and calculating gives:\[ v_A^2 = \frac{(3.95 - 0.0400 \times 9.81) \times 0.500}{0.0400} \]Calculate \( v_A^2 \) and then take the square root to find \( v_A \).
04

Calculate Kinetic Energy at Point B

At point \( B \), also consider centripetal motion. The equation is:\[ F_{N_B} + m \cdot g = \frac{m \cdot v_B^2}{R} \]Solving it for \( v_B \):\[ 0.680 + 0.0400 \times 9.81 = \frac{0.0400v_B^2}{0.500} \]Solve to find \( v_B \).
05

Calculate Velocity at Point B

Rearrange and solve for \( v_B^2 \):\[ v_B^2 = \frac{(0.680 + 0.0400 \times 9.81) \times 0.500}{0.0400} \]Determine \( v_B \) by taking the square root of \( v_B^2 \).
06

Calculate Change in Kinetic Energy

Using the velocities obtained, calculate the initial and final kinetic energies:\[ KE_A = \frac{1}{2} m v_A^2 \]\[ KE_B = \frac{1}{2} m v_B^2 \]Find the change in kinetic energy, \( \Delta KE = KE_B - KE_A \).
07

Calculate Work Done by Friction

Apply the work-energy principle:\[ W_{friction} = \Delta KE + m \cdot g \cdot (2R) \]Here, \( 2R \) is the change in height (since the block moves from bottom to top of the circle) and the gravitational potential energy change must be considered.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Force
Understanding the concept of normal force is important in scenarios involving circular motion. Normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the surface is the track of the circular path. When a block moves in a vertical circle, the normal force varies based on its position in the path.
At the bottom of the path (point A), the normal force is higher because it has to support not only the weight of the block but also provide the centripetal force required to keep it moving in a circle. The equation at point A is:
  • \( F_{N_A} = m imes g + \frac{m imes v_A^2}{R} \)
At the top of the path (point B), the normal force decreases. Here, gravity and the velocity provide the necessary centripetal force, reducing the normal force:
  • \( F_{N_B} + m \cdot g = \frac{m \cdot v_B^2}{R} \)
Normal force decreases at the top because gravity also acts in the direction needed for circular motion.
Kinetic Energy
Kinetic energy is a type of energy an object possesses due to its motion. In a circular motion scenario, the kinetic energy of an object is influenced by its speed. The formulas for kinetic energy at different points in the motion (A and B in this exercise) are given as:
  • At point A: \( KE_A = \frac{1}{2} m v_A^2 \)
  • At point B: \( KE_B = \frac{1}{2} m v_B^2 \)
The key idea here is that as the block moves from the bottom to the top of the circle, its velocity changes, and so does its kinetic energy. The change in velocity is due to the work done by forces during its motion. It's essential to calculate the velocities at both these points to determine their respective kinetic energies, reflecting the object's ability to perform work.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics. It relates the work done by forces acting on an object to the change in its kinetic energy. For a block moving in a vertical circle, as in this exercise, the principle can be used to calculate the work done by friction.After determining the kinetic energy at points A and B, the change in kinetic energy \( \Delta KE \) can be found. The formula for the work done by friction is:
  • \( W_{friction} = \Delta KE + m \cdot g \cdot (2R) \)
Here, \( m \cdot g \cdot (2R) \) represents the gravitational potential energy change, accounting for the height difference of \( 2R \) as the block moves from point A (bottom) to point B (top). The work-energy principle thus provides a comprehensive approach to finding the work done, by including changes in both kinetic and potential energy within the system.

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Most popular questions from this chapter

EP A 10.0 -kg microwave oven is pushed 8.00 m up the sloping surface of a loading ramp inclined at an angle of \(36.9^{\circ}\) above the horizontal, by a constant force \(\vec{\boldsymbol{F}}\) with a magnitude 110 \(\mathrm{N}\) and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250 . (a) What is the work done on the oven by the force \(F ?\) (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy. (e) Use \(\Sigma \vec{F}=m \vec{a}\) to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 8.00 \(\mathrm{m}\) . From this, compute the increase in the oven's kinetic energy, and compare it to the answer you got in part (d).

A force of 800 \(\mathrm{N}\) stretches a certain spring a distance of 0.200 \(\mathrm{m}\) . (a) What is the potential energy of the spring when it is stretched 0.200 \(\mathrm{m} ?\) (b) What is its potential energy when it is compressed 5.00 \(\mathrm{cm} ?\)

You are testing a new amusement park roller coaster with an empty car of mass 120 \(\mathrm{kg}\) . One part of the track is a vertical loop with radius 12.0 \(\mathrm{m}\) . At the bottom of the loop (point \(A )\) the car has speed \(25.0 \mathrm{m} / \mathrm{s},\) and at the top of the loop (point \(B )\) it has speed 8.0 \(\mathrm{m} / \mathrm{s} .\) As the car rolls from point \(A\) to point \(B\) , how much work is done by friction?

Down the Pole. A fireman of mass \(m\) slides a distance \(d\) down a pole. He sturts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance \(h \leq d\) above the ground and descended with negligible air resistance. (a) What average friction force did the fireman exert on the pole? Does your answer make sense in the special cases of \(h=d\) and \(h=0 ?\) (b) Find a numerical value for the average friction force a 75 -kg fireman exerts, for \(d=2.5 \mathrm{m}\) and \(h=1.0 \mathrm{m}\) (c) In terms of \(g, h,\) and \(d,\) what is the speed of the fireman when he is a distance \(y\) above the bottom of the pole?

CP A 120 -kg mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. A postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

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