/*! 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 43 At a waterpark, sleds with rider... [FREE SOLUTION] | 91Ó°ÊÓ

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At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring with force constant \(k=40.0 \mathrm{N} / \mathrm{cm}\) and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 \(\mathrm{kg}\) are pushed against the other end, compressing the spring 0.375 \(\mathrm{m}\) . The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 \(\mathrm{m} ?\)

Short Answer

Expert verified
(a) Speed = 2.834 m/s, (b) Speed = 2.397 m/s

Step by step solution

01

Convert Units for Spring Constant

First, we need to convert the spring constant from \(\mathrm{N/cm}\) to \(\mathrm{N/m}\). Given that \(k = 40.0\ \mathrm{N/cm}\), we convert this to \(\mathrm{N/m}\) as follows: \(k = 40.0 \, \mathrm{N/cm} \, \times\, 100 \, \mathrm{cm/m} = 4000 \, \mathrm{N/m}\).
02

Energy Conservation Principle

The principle of conservation of energy states that the initial mechanical energy (in the form of spring potential energy) will equal the final kinetic energy of the sled. The equation for spring potential energy is \(U = \frac{1}{2} k x^2\), and the equation for kinetic energy is \(K = \frac{1}{2} mv^2\).
03

Calculate Speed at Uncompressed Length

For part (a), when the spring returns to its uncompressed length, - Initial spring potential energy: \(U = \frac{1}{2} \times 4000 \, \mathrm{N/m} \times (0.375\, \mathrm{m})^2 = 281.25 \, \mathrm{J}\).- Final kinetic energy equal to the initial potential energy: \(K = \frac{1}{2} mv^2 = 281.25 \, \mathrm{J}\).- Solving for speed: \(v = \sqrt{\frac{2 \times 281.25}{70}} = \sqrt{8.036} = 2.834 \, \mathrm{m/s}\).
04

Calculate Speed When Compressed 0.200 m

For part (b), the spring is still compressed by 0.200 m.- Initial position energy: \(U_i = \frac{1}{2} \times 4000 \, \mathrm{N/m} \times (0.375 \, \mathrm{m})^2 = 281.25 \, \mathrm{J}\).- Energy at 0.200 m compressed: \(U_f = \frac{1}{2} \times 4000 \, \mathrm{N/m} \times (0.200 \, \mathrm{m})^2 = 80.0 \, \mathrm{J}\).- Kinetic energy when compressed 0.200 m: \(K = U_i - U_f = 281.25 \, \mathrm{J} - 80.0 \, \mathrm{J} = 201.25 \, \mathrm{J}\).- Solving for speed: \(v = \sqrt{\frac{2 \times 201.25}{70}} = \sqrt{5.75} = 2.397 \, \mathrm{m/s}\).

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Key Concepts

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

Understanding Spring Potential Energy
Spring potential energy is a type of energy stored in a compressed or stretched spring. It can be calculated using the formula:
\[ U = \frac{1}{2} k x^2 \]where:
  • \( U \) is the spring potential energy (in Joules),
  • \( k \) is the spring constant (in \( \mathrm{N/m} \)), a measure of the stiffness of the spring,
  • \( x \) is the displacement from the spring's equilibrium position (in meters).
Let's consider the given problem: the spring initially compresses by 0.375 meters. By substituting \( k = 4000 \, \mathrm{N/m} \) and \( x = 0.375 \, \mathrm{m} \), we find that the initial spring potential energy is a significant value that will eventually help propel the sled forward. It's crucial to understand that once the spring releases, this potential energy transforms into kinetic energy. This is the essence of the energy conservation we're using here, ensuring that no energy is lost in a frictionless scenario.
Exploring Kinetic Energy
Kinetic energy represents the energy that an object possesses due to its motion. The formula to calculate kinetic energy is:
\[ K = \frac{1}{2} mv^2 \]where:
  • \( K \) is the kinetic energy (in Joules),
  • \( m \) is the mass of the object (in kilograms),
  • \( v \) is the velocity of the object (in meters per second).
In this particular exercise, the sled starts from rest, meaning its initial kinetic energy is zero. As the spring returns to its uncompressed state, the potential energy stored in the spring is converted entirely into kinetic energy. Hence, the sled gains speed. For part (a), we observe that the energy conversion is complete at the full release of the spring, resulting in the sled's highest speed for this configuration. Part (b) shows that when the spring is still partially compressed, some potential energy remains, leading to a lower velocity of the sled compared to complete decompression.
The Role of a Frictionless Surface
In the context of this problem, the sled moves on a frictionless surface. This assumption simplifies calculations because it eliminates energy losses due to friction—a force that opposes motion and dissipates energy as heat. On a frictionless surface:
  • The total mechanical energy remains constant, assuming no external work is done.
  • All potential energy from the spring is transformed into kinetic energy without losses.
  • It allows us to apply the principle of conservation of energy directly without considering non-conservative forces.
The frictionless surface thus plays a key role by enabling the straightforward conversion of spring potential energy into kinetic energy. This ensures that the sled gains maximum possible speed as it is propelled across the surface. It serves an ideal scenario to apply theoretical principles of physics such as energy conservation effectively. This makes solving such exercises more manageable and provides a clear demonstration of these fundamental physical laws.

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Most popular questions from this chapter

Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) whileyou remain at rest. What is the average power supplied by the force you applied?

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BIO Chin-Ups, While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m} .\) (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-\mathrm{kg}\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

CALC A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x=0\) to \(x=6.9 \mathrm{m}\) as you apply a force with \(x\) -component \(F_{x}=-[20.0 \mathrm{N}+(3.0 \mathrm{N} / \mathrm{m}) x] .\) How much work does the force you apply do on the cow during this displacement?

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