/*! 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 42 A \(2.00-\mathrm{kg}\) block is ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 42

A \(2.00-\mathrm{kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) , compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. 7.30\()\) . (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

Short Answer

Expert verified
(a) The block's speed is 3.11 m/s. (b) It travels 1.64 m up the incline.

Step by step solution

01

Determine the Potential Energy in the Spring

The potential energy stored in the spring when compressed can be calculated using the formula for elastic potential energy: \[ PE_{spring} = \frac{1}{2}k x^2 \] Given values are: \( k = 400 \text{ N/m} \) and \( x = 0.220 \text{ m} \). Substitute these values into the equation: \[ PE_{spring} = \frac{1}{2} \times 400 \times (0.220)^2 \] Calculate the result, \( PE_{spring} = 9.68 \text{ J} \).
02

Convert Spring Potential Energy to Kinetic Energy

Once the block is released from the spring, all potential energy is converted into kinetic energy because the surface is frictionless. The kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \] where \( m = 2.00 \text{ kg} \) is the mass of the block, and \( v \) is its velocity. Set the kinetic energy equal to the spring potential energy: \[ 9.68 = \frac{1}{2} \times 2.00 \times v^2 \] Solve for \( v \): \[ v^2 = \frac{9.68}{1.00} = 9.68 \] \[ v = \sqrt{9.68} = 3.11 \text{ m/s} \].
03

Calculate the Height Reached on the Incline

As the block slides up the incline, kinetic energy is converted to gravitational potential energy given by \( PE_{gravity} = mgh \). Set \( KE = PE_{gravity} \): \[ \frac{1}{2} mv^2 = mgh \] The mass \( m \) cancels out: \[ \frac{1}{2} (3.11)^2 = 9.8h \] Solve for \( h \): \[ h = \frac{9.68}{9.8} = 0.987 \text{ m} \].
04

Find the Distance Traveled Along the Incline

To find the distance \( d \) traveled along the incline, use the relationship between height and distance on an incline: \[ h = d \sin(37^{\circ}) \] Substitute the value of \( h \): \[ 0.987 = d \times \sin(37^{\circ}) \] \[ d = \frac{0.987}{\sin(37^{\circ})} \approx \frac{0.987}{0.6018} \approx 1.64 \text{ m} \].

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Key Concepts

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

Elastic Potential Energy
When a spring is compressed, it stores energy known as elastic potential energy. This type of potential energy can be calculated using the formula:
  • \( PE_{spring} = \frac{1}{2} k x^2 \)
where \( k \) is the spring constant, indicating the stiffness of the spring, and \( x \) is the displacement or compression from its equilibrium position.
In our exercise, a spring with a constant \( k \) of 400 N/m is compressed by 0.220 m. Substituting these values, we find that the elastic potential energy stored is approximately 9.68 Joules. This energy will be converted into kinetic energy once the spring is released, assuming no energy loss due to friction. Understanding how springs store and release energy helps explain many real-world mechanical systems.
Kinetic Energy
Kinetic energy is the energy of motion. When our block is released from the spring, the stored elastic potential energy in the spring is converted into kinetic energy.
Kinetic energy is given by:
  • \( KE = \frac{1}{2} mv^2 \)
where \( m \) is the mass of the object and \( v \) its velocity.
In our case of a 2.00 kg block, the spring's entire potential energy changes into kinetic energy because the surface is frictionless. Solving \( \frac{1}{2} \times 2.00 \times v^2 = 9.68 \) gives a velocity of approximately 3.11 m/s. Understanding kinetic energy is crucial as it informs how objects move and interact in the absence of external forces.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, used as a simple machine. It reduces the effort needed to elevate objects by extending the distance over which the force is applied.
In our problem, the incline's slope is given as \(37.0^{\circ}\). As the block moves on this incline, kinetic energy is converted into gravitational potential energy, slowing down the block until it momentarily stops.
The relationship between the height \( h \) and the distance \( d \) traveled on the incline uses trigonometry:
  • \( h = d \cdot \sin(\theta) \)
where \( \theta \) is the angle of the incline. In our exercise, the block rises 0.987 m up the incline, resulting in a traveled distance of approximately 1.64 m. Understanding the dynamics of inclined planes is essential in solving various physics problems involving slopes and ramps.
Gravitational Potential Energy
As the block moves up the incline, its kinetic energy is turned into gravitational potential energy. This form of energy depends on the object's height above a reference level and is calculated using:
  • \( PE_{gravity} = mgh \)
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
For the moving block, as kinetic energy equals gravitational potential energy when at the peak height, we find \( h \) by using the block's initial kinetic energy of 9.68 Joules: \( h = \frac{9.68}{9.8} \), resulting in approximately 0.987 m.
This principle is foundational in physics, allowing us to predict how high an object will rise given its kinetic energy.

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Most popular questions from this chapter

A \(60.0-\mathrm{kg}\) skier starts from rest at the top of a ski slope 65.0 \(\mathrm{m}\) high. (a) If frictional forces do \(-10.5 \mathbf{k J}\) of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow, where \(\mu_{\mathrm{k}}=0.20\) If the patch is 82.0 \(\mathrm{m}\) wide and the average force of air resistance on the skier is 160 \(\mathrm{N}\) , how fast is she going after crossing the patch? (c) The skier hits a snowdrift and Penetrates 2.5 \(\mathrm{m}\) into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

A 10.0 -kg microwave oven is pushed 8.00 \(\mathrm{m}\) up the sloping surface of a loading ramp inclined at an angle of \(36.9^{\circ}\) above the horizontal, by a constant force \(\vec{F}\) 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 \(\vec{F} ?\) (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 \(\Sigma \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}\) 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 crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: (a) Take the level at which the potential energy is zero to be at the bottom of the ramp with \(y\) positive upward. (b) Take the zero level for potential energy to be at the top of the ramp with y positive upward. (c) Why did the normal force not enter into your solution?

In a truck loading station at a post office, a small 0.200 \(\mathrm{kg}\) package is released from rest at point \(A\) on a track that is one-quarter of a circle with radius 1.60 \(\mathrm{m}(\mathrm{Fig} .7 .39) .\) The size of the package is much less than \(1.60 \mathrm{m},\) so the package can be treated as a particle. It slides down the track and reaches point \(B\) with a speed of 4.80 \(\mathrm{m} / \mathrm{s}\) . From point \(B,\) it slides on a level surface a distance of 3.00 \(\mathrm{m}\) to point \(C,\) where it comes to rest (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much work is done on the package by friction as it slides down the circular are from \(A\) to \(B ?\)

A \(120-\mathrm{kg}\) mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. \(\mathrm{A}\) 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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