/*! 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 -kg block is pushed again... [FREE SOLUTION] | 91Ó°ÊÓ

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A 2.00 -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" (Fig. \(\mathrm{P7} .42\) ). (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 speed of the block is 3.10 m/s. (b) The block travels 1.55 m up the incline.

Step by step solution

01

Calculate Potential Energy in the Spring

First, determine the potential energy stored in the compressed spring using the formula for spring potential energy \( U_s = \frac{1}{2} k x^2 \), where \( k = 400 \, \text{N/m} \) and \( x = 0.220 \, \text{m} \). Substitute in the given values: \[ U_s = \frac{1}{2} \times 400 \, \text{N/m} \times (0.220 \, \text{m})^2. \]Calculate to find \( U_s. \)
02

Convert Spring's Potential Energy to Kinetic Energy

All the potential energy stored in the spring is converted into kinetic energy \( K \) when the block is released. The kinetic energy of the block is given by \( K = \frac{1}{2} mv^2 \), where \( m = 2.00 \, \text{kg} \) and \( v \) is the speed we want to find. Set this equal to the potential energy from the spring: \[ \frac{1}{2} mv^2 = U_s. \]Solve for \( v \) to find the block's speed on the horizontal surface.
03

Determine the Speed of the Block

Solve the equation from Step 2: \[ \frac{1}{2} \times 2.00 \, \text{kg} \times v^2 = 9.68 \, \text{J}. \]This simplifies to \[ v^2 = \frac{9.68}{1.00} \, \text{m}^2/\text{s}^2.\]Calculate \( v \) to find the speed of the block.
04

Analyze Energy on the Incline

Once on the incline, the kinetic energy is converted into gravitational potential energy \( U_g \) as the block moves upward. \( U_g = mgh \), where \( h \) is the height. The potential energy is given by \[ mgh = \frac{1}{2} mv^2. \]From the geometry of the incline, height \( h \) can be written in terms of distance \( d \) moved up the incline: \( h = d \sin(37^\circ) \). Substitute \( h \) and solve for \( d \).
05

Calculate Distance Travelled Up the Incline

Substitute \( h = d \sin(37^\circ) \) into \[ U_g = mgh, \] which means \[ m \cdot g \cdot d \cdot \sin(37^\circ) = \frac{1}{2} \cdot 2.00 \cdot v^2. \]Simplify and solve for \( d \) knowing \( g = 9.8 \text{ m/s}^2 \).

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Key Concepts

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

Spring Potential Energy
When a block is compressed against a spring, it stores energy called spring potential energy. This energy is determined by the formula \( U_s = \frac{1}{2} k x^2 \). Here, \( k \) is the spring constant and \( x \) is the distance the spring is compressed or stretched. The spring constant is a measure of the spring's stiffness, and a higher \( k \) means a stiffer spring.
The term \( \frac{1}{2} \) in the formula represents the distribution of force as the spring is compressed or released. As the block pushes against the spring to a compression of 0.220 m, the spring potential energy becomes stored, ready to transform into another form of energy when the spring is released.
Kinetic Energy Conversion
Kinetic energy is the energy an object possesses due to its motion. When the spring is released, the potential energy stored in it is converted into kinetic energy, propelling the block forward. The kinetic energy \( K \) can be calculated using the formula \( K = \frac{1}{2} mv^2 \) where \( m \) is the mass of the block and \( v \) is its velocity.
This conversion process adheres to the law of energy conservation, meaning the total energy remains constant as it merely transfers from one form (potential) to another (kinetic). For the block, calculating its speed just involves equating the kinetic energy to the initial spring potential energy and solving for \( v \).
  • The initial potential energy \( U_s \) in the spring becomes the kinetic energy \( K \) of the block.
  • Use \( \frac{1}{2} mv^2 = U_s \) to find the velocity \( v \).
Gravitational Potential Energy
As the block travels up the incline, kinetic energy is converted into gravitational potential energy. This energy depends on the height function \( h \) and is calculated as \( U_g = mgh \). Here, \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height on the incline.
Since the block moves on a slope, we need to link the height \( h \) with the distance \( d \) traveled along the incline using the angle \( \theta \). We use \( h = d \sin(\theta) \), which allows us to substitute \( h \) in the potential energy equation. This part of the calculation helps determine how far the block moves up before its kinetic energy is completely converted to gravitational potential energy.
  • The higher the block moves, the greater the gravitational potential energy.
  • When all kinetic energy is converted, the block momentarily stops before sliding back.
Inclined Plane Mechanics
An inclined plane in physics is a flat surface tilted at an angle to the horizontal. Analyzing mechanics on an inclined plane involves understanding forces like gravity acting at an angle. When the block travels up the frictionless incline, we consider components of gravitational force along and perpendicular to the plane.
The distance \( d \) traveled on the incline directly relates to how high the block gets. With all input energy being conserved and no friction to dissipate it, the motion of the block simplifies dramatically. The cosine and sine components of gravitational force help us compute movement and potential energy, utilizing the inclination angle \( \theta \).
  • Solve for the incline distance using triangle relationships \( h = d \sin(\theta) \).
  • Energy conversion simplifies calculations due to lack of friction.
  • Understanding forces acting along the plane aids in comprehending motion.

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Most popular questions from this chapter

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