/*! 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 57 A block slides along a track fro... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 57

A block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance \(d\). The block's initial speed \(v_{0}\) is \(6.0 \mathrm{~m} / \mathrm{s}\), the height difference \(h\) is \(1.1 \mathrm{~m}\), and \(\mu_{k}\) is \(0.60\). Find \(d\).

Short Answer

Expert verified
The stopping distance \(d\) is calculated using energy conservation and work-energy principles.

Step by step solution

01

Calculate Initial Mechanical Energy

The block starts with kinetic energy on the lower level and potential energy from the higher level it ascends to. The initial mechanical energy is given by \( KE_i = \frac{1}{2} m v_0^2 \). Since there is no initial potential energy, total initial mechanical energy is the initial kinetic energy, \( ME_i = \frac{1}{2} m (6.0)^2 \).
02

Calculate Potential Energy at Higher Level

When the block reaches the higher level, it has gained potential energy due to the height difference \( h = 1.1 \, \text{m} \). The potential energy will be \( PE = mgh \).
03

Use Energy Conservation to Find Remaining Kinetic Energy

Apply the conservation of energy principle: initial mechanical energy equals potential energy at height plus kinetic energy, i.e., \( ME_i = PE + KE_f \). Hence, \( \frac{1}{2} m (v_0)^2 = mgh + \frac{1}{2} m v_f^2 \). Solve for \( v_f^2 \).
04

Calculate Work Done by Friction

The work done by the friction to stop the block is \( W = F_f d \). The frictional force \( F_f \) is given by \( F_f = \mu_k mg \), where \( \mu_k = 0.60 \). So, the work done \( W = \mu_k mgd \).
05

Relate Work Done to Kinetic Energy

The work done by friction should be equal to the remaining kinetic energy when the block starts at the higher level. Set \( \mu_k mgd = \frac{1}{2} m v_f^2 \) from Step 3. Simplify to solve for \( d \), the stopping distance.
06

Solve for Distance \(d\)

Remove \( m \) (mass) from the equation since it cancels out and solve for \( d \): \( d = \frac{v_f^2}{2 \mu_k g} \). Substitute known values and compute \( d \).

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics, expressing that energy cannot be created or destroyed in an isolated system. Instead, it can only be transformed from one form into another. For our block on the track:
  • It starts with kinetic energy due to its speed at the lower level.
  • As it goes up the valley to the higher level, its kinetic energy converts into potential energy.
  • Potential energy increases with height and is calculated as \( PE = mgh \). Here, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
This exercise shows how the block's initial mechanical energy (entirely kinetic at the start) shares as potential energy and any remaining kinetic energy at the top. In this case, the conservation of energy helps account for how the block can reach the top of the hill with its initial speed, converting energy without loss to friction until it is stopped farther up.
Frictional Force
Frictional force is an opposing force that acts parallel to the surfaces in contact, opposing motion. In the context of the exercise:
  • On the lower level, the track is frictionless, meaning the block moves without loss of mechanical energy to heat or sound.
  • Upon reaching the higher level, friction comes into play, working to halt the block. The kinetic frictional force \( F_f \) is calculated using \( F_f = \mu_k mg \).
  • \( \mu_k \) is the coefficient of kinetic friction; in this problem, it's given as 0.60, showing a moderate level of frictional resistance.
The work done by this friction, \( W = F_f d \), helps find how far the block can travel before stopping, showing the vital role of frictional forces in balancing energy within a mechanical system.
Kinetic Energy
Understanding kinetic energy is key to solving this exercise. It's the energy associated with the motion of an object, given by the formula \( KE = \frac{1}{2} mv^2 \). Here's how it plays out in our scenario:
  • The block starts with kinetic energy determined by its initial speed of 6.0 m/s. This is the only energy type at the starting level since the potential energy is initially zero.
  • As the block ascends, its kinetic energy decreases, giving way to potential energy at the higher level.
  • Even at the higher level, some kinetic energy remains, which is then countered by friction to bring the block to a stop.
Thus, kinetic energy not only determines the block's initial and any remaining speed but also becomes the source of the energy opposing friction. Understanding this helps trace how energy facilitates movement and confronts resistive forces.

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Most popular questions from this chapter

A \(60 \mathrm{~kg}\) skier leaves the end of a ski-jump ramp with a velocity of \(24 \mathrm{~m} / \mathrm{s}\) directed \(25^{\circ}\) above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of \(22 \mathrm{~m} / \mathrm{s}\), landing \(14 \mathrm{~m}\) vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

A collie drags its bed box across a floor by applying a horizontal force of \(8.0 \mathrm{~N}\). The kinetic frictional force acting on the box has magnitude \(5.0 \mathrm{~N}\). As the box is dragged through \(0.70 \mathrm{~m}\) along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

When a click beetle is upside down on its back, it jumps upward by suddenly arching its back, transferring energy stored in a muscle to mechanical energy. This launching mechanism produces an audible click, giving the beetle its name. Videotape of a certain click-beetle jump shows that a beetle of mass \(m=4.0 \times 10^{-6} \mathrm{~kg}\) moved directly upward by \(0.77 \mathrm{~mm}\) during the launch and then to a maximum height of \(h=0.30 \mathrm{~m}\). During the launch, what are the average magnitudes of (a) the external force on the beetle's back from the floor and (b) the acceleration of the beetle in terms of \(g\) ?

A \(9.40 \mathrm{~kg}\) projectile is fired vertically upward. Air drag decreases the mechanical energy of the projectile-Earth system by \(68.0 \mathrm{~kJ}\) during the projectile's ascent. How much higher would the projectile have gone were air drag negligible?

Ice flake is released from the edge of a hemispherical bowl whose radius \(r\) is \(22.0\) \(\mathrm{cm}\). The flake-bowl contact is frictionless. (a) How much work is done on the flake by the gravitational force during the flake's descent to the bottom of the bowl? (b) What is the change in the potential energy of the flake-Earth system during that descent? (c) If that potential energy is taken to be zero at the bottom of the bowl, what is its value when the flake is released? (d) If, instead, the potential energy is taken to be zero at the release point, what is its value when the flake reaches the bottom of the bowl? (e) If the mass of the flake were doubled, would the magnitudes of the answers to (a) through (d) increase, decrease, or remain the same?
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