/*! 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 52 A block with mass \(m=\) \(\begi... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 52

A block with mass \(m=\) \(\begin{array}{lll}0.200 \mathrm{~kg} & \text { is placed against a com- }\end{array}\) pressed spring at the bottom of a ramp that is at an angle of \(53.0^{\circ}\) above the horizontal. The spring has \(8.00 \mathrm{~J}\) of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of \(3.00 \mathrm{~m},\) its speed is \(4.00 \mathrm{~m} / \mathrm{s} .\) What is the magnitude of the friction force that the ramp exerts on the block while the block is moving?

Short Answer

Expert verified
The magnitude of the friction force that the ramp exerts on the block while the block is moving is \(1.616 \, N\).

Step by step solution

01

Establish the principle of conservation of energy

According to the principle of energy conservation, the total mechanical energy at the start of the motion (just when the spring is released) should be equal to the total mechanical energy after the block has moved up the ramp, taking into account the work done against friction. At the start of the motion, the total mechanical energy is the potential energy of the spring, which is \(8.00 \, J\). After the block has moved up the ramp, the mechanical energy is the sum of the block鈥檚 kinetic energy and the gravitational potential energy at a height \(h\) from the bottom of the ramp. The kinetic energy of the block is given by \(0.5 \cdot m \cdot v^2 = 0.5 \cdot 0.200 \, kg \cdot (4.00 \, m/s)^2 = 0.800 \, J\). The gravitational potential energy at a height \(h\) is given by \(m \cdot g \cdot h\), where \(g = 9.8 \, m/s^2\) is the acceleration due to gravity and \(h = 3.00 \, m \cdot \sin{53.0}\) is the height the block has moved upward along the ramp. Thus, \(m \cdot g \cdot h = 0.200 \, kg \cdot 9.8 \, m/s^2 \cdot 3.00 \, m \cdot \sin{53.0} = 2.352 \, J\). The work done against friction is equal to the initial mechanical energy minus the final mechanical energy, and it is negative since it's done against the direction of motion.
02

Compute the work done against friction

The work done (\(W\)) against friction is given by \(W = total \, mechanical \, energy \, initial - total \, mechanical \, energy \, final\). Plugging the values calculated, \(W = 8.00 \, J - (0.800 \, J + 2.352 \, J) = 8.00 \, J - 3.152 \, J = 4.848 \, J\).
03

Apply the work-energy theorem

The work-energy theorem relates the work done on an object to the change in its kinetic energy. In our case, the work done against the frictional force is equivalent to the frictional force multiplied by the block's displacement along the ramp: \(W = F_{friction} \times d\), where \(d = 3.00 \, m\) (the distance the block has moved up the ramp). Solving for \(F_{friction}\) gives us \(F_{friction} = \frac{W}{d} = \frac{4.848 \, J}{3.00 \, m} = 1.616 \, N\).

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Key Concepts

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

Frictional Force
Frictional force is a fundamental concept in physics relating to how objects interact with surfaces. When the block slides up the ramp, it encounters a force opposing its motion called frictional force. This force arises when two surfaces, in this case, the block and the ramp, are in contact and move relative to each other. The frictional force ultimately depends on two main factors:
  • The nature of the surfaces, often described by the coefficient of friction.
  • The normal force, which, in this scenario, is affected by the block鈥檚 weight and the incline of the ramp.
Friction transforms useful mechanical energy into thermal energy. In our exercise, the frictional force is what dampens the total mechanical energy, which initially is stored as elastic potential energy of the spring. After the spring releases, part of this energy is dissipated by the frictional force as the block moves up the ramp. It is crucial to understand that friction is necessary for controlling movement, even though it can sometimes be seen as a hindrance.
Elastic Potential Energy
Elastic potential energy is the energy stored in an object when it is deformed, such as stretching or compressing a spring. In our problem, the spring has an elastic potential energy of 8.00 J. This energy is stored as a result of compressing the spring, and it can do work when released.
  • Elastic potential energy is determined by the equation: \[ U_e = \frac{1}{2} k x^2 \]where \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position.
  • When the spring is released, this stored energy is converted into kinetic energy and gravitational potential energy.
For this exercise, the block converted the elastic potential energy from the spring into movement up the incline. This demonstrates how stored energy can be transformed and used to perform work on an object, leading to changes in motion.
Work-Energy Theorem
The work-energy theorem is a cornerstone of physics that relates the work done on an object to its change in kinetic energy. It's vital in analyzing forces acting over a distance and how they alter an object's motion.
  • According to the theorem, the work done by all forces acting on an object equals the change in the object's kinetic energy: \[ W = \Delta KE = KE_{final} - KE_{initial} \]
  • In this problem, the theorem helps calculate the frictional force by connecting the work done against it to the energy transformations occurring.
  • The total energy at the starting point minus the total energy after moving, gives us the work done against friction.
Utilizing the work-energy theorem, we determined how much of the initial potential energy was "lost" to friction as the block ascended the ramp. This approach allows for understanding energy distribution without directly measuring the forces involved. Through simple calculations, we unravel the interactions that occur, and the theorem serves as a tool to deduce otherwise complex energy exchanges.

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Most popular questions from this chapter

A conservative force \(\vec{F}\) is in the \(+x\) -direction and has magnitude \(F(x)=\alpha /\left(x+x_{0}\right)^{2},\) where \(\alpha=0.800 \mathrm{~N} \cdot \mathrm{m}^{2}\) and \(x_{0}=0.200 \mathrm{~m}\). (a) What is the potential- energy function \(U(x)\) for this force? Let \(U(x) \rightarrow 0\) as \(x \rightarrow \infty .\) (b) An object with mass \(m=0.500 \mathrm{~kg}\) is released from rest at \(x=0\) and moves in the \(+x\) -direction. If \(\vec{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x=0.400 \mathrm{~m} ?\)

A 1.20 kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{~N} / \mathrm{m}\) that is compressed \(15.0 \mathrm{~cm} .\) When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

CALC A small block with mass \(0.0400 \mathrm{~kg}\) is moving in the \(x y\) -plane. The net force on the block is described by the potential-energy function \(U(x, y)=\left(5.80 \mathrm{~J} / \mathrm{m}^{2}\right) x^{2}-\left(3.60 \mathrm{~J} / \mathrm{m}^{3}\right) \mathrm{y}^{3}\). What are the magnitude and direction of the acceleration of the block when it is at the point \((x=0.300 \mathrm{~m}, y=0.600 \mathrm{~m}) ?\)

You are an industrial engineer with a shipping company. As part of the package-handling system, a small box with mass \(1.60 \mathrm{~kg}\) is placed against a light spring that is compressed \(0.280 \mathrm{~m}\). The spring, whose other end is attached to a wall, has force constant \(k=45.0 \mathrm{~N} / \mathrm{m}\). The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is \(\mu_{k}=0.300 .\) When the box has traveled \(0.280 \mathrm{~m}\) and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?

A 90.0 kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 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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