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Chapter 4: Problem 61

A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of \(7.0 \mathrm{~m} / \mathrm{s}\). If the coefficient of friction between the sled's runners and the snow is \(0.050\) and the boy and sled together weigh \(600 \mathrm{~N}\), how far does the sled travel on the level surface before coming to rest?

Short Answer

Expert verified
The sled travels 50 meters on the level surface before coming to rest.

Step by step solution

01

Determine the friction force

The friction force can be calculated by multiplying the coefficient of friction by the weight. In this case, that would be \(0.050 \times 600 \mathrm{~N} = 30 \mathrm{~N}\).
02

Utilize the work-energy principle

According to the work-energy principle, the work done should be equal to the change in the kinetic energy. The work done by the friction force (which acts opposite to the direction of motion) is \(W = f \times d\), where \(f\) is the friction force and \(d\) is the distance to be determined. The initial kinetic energy of the sled (when it hits the level surface) is \(\frac{1}{2} m v^2\), where \(m\) is the mass and \(v\) is the speed of the sled. Given that the sled comes to rest eventually, its final kinetic energy is zero.
03

Calculate the mass of sled

The weight of the sled is equivalent to \(mass \times gravity\). Given that the weight is \(600 \mathrm{~N}\) and the acceleration due to gravity is approximately \(9.8 \mathrm{~m/s}^2\), the mass of the sled is \(\frac{600 \mathrm{~N}}{9.8 \mathrm{~m/s}^2} = 61.2 \mathrm{~kg}\).
04

Obtain the initial kinetic energy

The initial kinetic energy can be determined using \(\frac{1}{2} m v^2\), which amounts to \(\frac{1}{2} \times 61.2 \mathrm{~kg} \times (7.0 \mathrm{~m/s})^2 = 1500 \mathrm{~J}\).
05

Calculate the distance

From the work-energy principle, we have that \(f \times d = 1500 \mathrm{~J}\). Therefore, the distance the sled travels on the level surface before coming to rest is \(\frac{1500 \mathrm{~J}}{30 \mathrm{~N}} = 50 \mathrm{~m}\).

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Key Concepts

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

Friction Force
Friction is a force that opposes the relative motion between two surfaces in contact. In this scenario, the friction force is what gradually slows down the sled as it travels on the level surface. Here's how it works:
  • The sled's runners are in contact with the snow, causing frictional resistance.
  • The strength of this frictional force depends on two things: the coefficient of friction and the weight of the sled.
  • In our exercise, the coefficient of friction is 0.050.
  • The sled and boy weigh 600 N. From this, we calculate the friction force as: \[ \text{Friction force} = \text{Coefficient of friction} \times \text{Weight} = 0.050 \times 600 \mathrm{~N} = 30 \mathrm{~N} \]
This friction force is constantly working against the sled's motion, eventually bringing it to a stop.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For the sled:
  • The sled has kinetic energy because it is moving.
  • It starts with a speed of 7.0 m/s on a level surface.
  • Kinetic energy can be calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass and \( v \) is the velocity.
  • To find the initial kinetic energy:
    • The sled's weight is 600 N, which gives us a mass of 61.2 kg (since weight = mass \( \times \) gravity).
    • Thus, the initial kinetic energy is: \[ \frac{1}{2} \times 61.2 \mathrm{~kg} \times (7.0 \mathrm{~m/s})^2 = 1500 \mathrm{~J} \]
When the sled stops moving, all of its kinetic energy has been used up.
Motion
Motion refers to the change in position of an object over time. In physics problems like this, motion is usually influenced by forces like friction. Here鈥檚 a breakdown:
  • The sled is moving at the bottom of the hill with a speed of 7.0 m/s.
  • As it travels across the leveled snow, the friction force from the snow opposes its motion.
  • This opposition causes the sled to eventually come to a rest.
Think of motion as a story:
  • The sled starts with a "high-speed chapter" when it hits the flat ground."
  • Friction slowly writes the "slowing down chapter," consuming the sled's kinetic energy.
  • The story ends when the sled comes to a stop.
Understanding motion is crucial in predicting how and when objects will stop when forces like friction are involved.
Physics Problem-Solving
Solving physics problems often involves putting together several concepts. Here's a simplified approach based on our sled and snow problem:
  • Identify Given Information: Start by noting down what's given - coefficient of friction, sled's weight, and initial speed.
  • Find Relevant Formulas: Use the friction force formula and kinetic energy formula to connect motion with energy changes.
  • Relate Work to Energy: The work-energy principle is key. It equates the work done by forces to changes in the energy of an object.
  • Calculate Each Step:
    • First, calculate the friction force.
    • Next, determine the sled's kinetic energy.
    • Finally, use these to find out how far the sled travels before stopping (i.e., solve for distance in the equation \( f \times d = \text{Initial Kinetic Energy} \)).
A well-structured approach simplifies even the trickiest problems by breaking them down into manageable pieces. The more you practice, the more intuitive these steps become in your problem-solving toolkit.

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Most popular questions from this chapter

A sled weighing \(60.0 \mathrm{~N}\) is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is \(0.100, \overline{\mathrm{A}}\) penguin weighing \(70.0 \mathrm{~N}\) rides on the sled, as in Figtire \(\mathrm{P} 4.78\). If the coefficient of static friction between penguin and sled is \(0.700\), find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.

A coin is placed near one edge of a book lying on a table, and that edge of the book is lifted until the coin just slips down the incline as shown in Figure \(\mathrm{P} 4,70\). The angle of the incline, \(\theta_{c}\), called the critical angle, is measured. (a) Draw a frec-body diagram for the coin when it is on the verge of slipping and identify all forces acting on it. Your free- body diagram should include a force of static friction acting up the incline. (b) Is the magnitude of the friction force equal to \(\mu_{3} n\) for angles less than \(\theta_{c}\) ? Explain. What can you definitely say about the magnitude of the friction force for any angle \(\theta \leq \theta_{c} ?\) (c) Show that the coefficient of static friction is given by \(\mu_{\mathrm{s}}=\tan \theta_{c}\). (d) Once the coin starts to slide down the incline, the angle can be adjusted to a new value \(\theta_{c}^{\prime} \leq \theta_{c}\) such that the coins moves down the incline with constant speed. How does observation enable you to obtain the coefficient of kinetic friction?

\(A\) freight train has a mass of \(1.5 \times 10^{7} \mathrm{~kg}\). If the locomotive can exert a constant pull of \(7.5 \times 10^{5} \mathrm{~N}\), how long does it take to increase the speed of the train from rest to \(80 \mathrm{~km} / \mathrm{h} ?\)

A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of \(1.2 \mathrm{~m} / \mathrm{s}\) in \(0.80 \mathrm{~s} .\) The elevator travels with this constant speed for \(5.0 \mathrm{~s}\), undergoes a uniform negative acceleration for \(1.5 \mathrm{~s}\), and then comes to rest. What does the spring scale register (a) before the elevator starts to move? (b) During the first \(0.80 \mathrm{~s}\) of the elevator's ascent? (c) While the elevator is traveling at constant speed? (d) During the elevator's negative acceleration?

A chinook salmon has a maximum underwater speed of \(3.0 \mathrm{~m} / \mathrm{s}\), and can jump out of the water vertically with a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). A record salmon has a length of \(1.5 \mathrm{~m}\) and a mass of \(61 \mathrm{~kg}\). When swimming upward at constant speed, and neglecting buoyancy, the fish experiences three forces: an upward force \(F\) exerted by the tail fin, the downward drag force of the water, and the downward force of gravity. As the fish leaves the surface of the water, however, it experiences a net upward force causing it to accelerate from \(3.0 \mathrm{~m} / \mathrm{s}\) to \(6.0 \mathrm{~m} / \mathrm{s}\). Assuming the drag force disappears as soon as the head of the fish breaks the surface and \(F\) is exerted until two-thirds of the fish's length has left the water, determine the magnitude of \(F\).
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