/*! 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 49 A \(25 \mathrm{~kg}\) bear slide... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

Short Answer

Expert verified
(a) -2940 J, (b) 392 J, (c) 277.67 N.

Step by step solution

01

Calculate Gravitational Potential Energy Change

We begin by finding the change in gravitational potential energy (\(\Delta U\)) of the bear-Earth system using the formula:\[\Delta U = m g h\]where \(m = 25 \, \mathrm{kg}\) is the mass of the bear, \(g = 9.8 \, \mathrm{m/s^2}\) is the acceleration due to gravity, and \(h = -12 \, \mathrm{m}\) since the bear slides down. Thus,\[\Delta U = 25 \times 9.8 \times (-12) = -2940 \, \mathrm{J}\]The negative sign indicates that the gravitational potential energy decreases as the bear moves down.
02

Calculate Kinetic Energy Before Impact

The kinetic energy of the bear just before hitting the ground can be calculated using the kinetic energy formula:\[KE = \frac{1}{2} m v^2\]where \(v = 5.6 \, \mathrm{m/s}\) is the final speed of the bear. Substituting the known values:\[KE = \frac{1}{2} \times 25 \times (5.6)^2 = 392 \, \mathrm{J}\]This value represents the kinetic energy of the bear just before it reaches the ground.
03

Calculate Frictional Force

To find the average frictional force, we use the work-energy principle which states the work done by all forces is the change in kinetic energy (\Delta KE) plus the change in potential energy (\Delta U).\[\Delta KE = KE - 0 = 392 \, \mathrm{J}\]The work done by friction (\(W_f\)) can be computed as:\[W_f = \Delta KE - \Delta U = 392 + 2940 = 3332 \, \mathrm{J}\]The work done by friction is also given by \(-f_d \times d\) where \(f_d\) is the frictional force and \(d = 12 \, \mathrm{m}\) is the distance.\[-f_d \times 12 = 3332 \ f_d = \frac{3332}{12} = 277.67 \, \mathrm{N}\]Thus, the average frictional force acting on the sliding bear is approximately \(277.67 \, \mathrm{N}\).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position relative to the Earth. It is an essential concept in mechanics and plays a huge role in understanding the energy changes during the movement of objects. When an object like a bear slides down a tree, its height above the ground decreases. This results in a change in gravitational potential energy. The formula to compute this change is given by the equation \(\Delta U = mgh\). Here, \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately \(9.8 \, \mathrm{m/s^2}\) on Earth), and \(h\) is the change in height.
  • As the bear moves down, \(h\) becomes negative, indicating a loss in potential energy.
  • The change in potential energy is the difference between the initial and final energy values.
  • In this particular problem, the gravitational potential energy decreases by \(-2940 \, \mathrm{J}\).
Understanding this process helps in predicting how potential energy transforms into other forms like kinetic energy as objects move.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is directly proportional to the mass of the object and the square of its velocity. For any moving object, such as the bear in the problem, the kinetic energy just before hitting the ground can be calculated using the formula \(KE = \frac{1}{2} mv^2\). This formula tells us that:
  • The mass \(m\) significantly affects the amount of kinetic energy, even more so than its velocity.
  • Velocity \(v\) has a greater influence since it is squared in the formula.
In our example, where the bear is sliding down with a speed of \(5.6 \, \mathrm{m/s}\) just before hitting the ground, its kinetic energy is calculated to be \(392 \, \mathrm{J}\). This kinetic energy represents the energy transformation from gravitational potential energy due to its movement downwards, depicting how energy manifests in different forms during motion.
Frictional Force
Frictional force is a force that opposes motion between two surfaces in contact. It is an essential concept in mechanics that affects how objects move or stop when in contact with other surfaces. In the situation of the bear sliding down a tree, the frictional force acts upwards, opposing its downward motion.
  • The average frictional force can be determined using the work-energy principle. \(-f_d \times d\) represents the work done by the frictional force over a distance \(d\).
  • In this case, the calculated frictional force is approximately \(277.67 \, \mathrm{N}\) which shows how much force was needed to counteract part of the bear's movement energy.
  • This force helps slow down the bear, converting some of the sliding kinetic energy into heat and sound.
Understanding frictional force allows us to get a clearer picture of how energy is distributed in systems and how objects interact with their environments, playing an essential role in stopping or slowing the motion.

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Most popular questions from this chapter

The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

A conservative force \(\vec{F}=(6.0 x-12) \hat{\mathrm{i}} \mathrm{N}\) where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\), with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of \(x\) is the potential energy equal to zero?

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A \(75 \mathrm{~g}\) Frisbee is thrown from a point \(1.1 \mathrm{~m}\) above the ground with a speed of \(12 \mathrm{~m} / \mathrm{s}\). When it has reached a height of \(2.1 \mathrm{~m}\), its speed is \(10.5 \mathrm{~m} / \mathrm{s}\). What was the reduction in \(E_{\mathrm{mec}}\) of the Frisbee-Earth system because of air drag?

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) above a vertical spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{0} ?(\mathrm{~d})\) If the block were released from height \(2.00 h_{0}\) above the spring, what would be the maximum compression of the spring?

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