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Physics / Physics with MasteringPhysics 4 / Chapter 13 / Problem 55P

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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Solved: A 0.505-kg block slides on a frictionless

Chapter 13, Problem 55P

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QUESTION:

Problem 54P

A 0.505-kg block slides on a frictionless horizontal surface with a speed of 1.18 m/s. The block encounters an unstretched spring and compresses it 23.2 cm before coming to rest. (a) What is the force constant of this spring? (b) For what length of time is the block in contact with the spring before it comes to rest? (c) If the force constant of the spring is increased, docs the time required to stop the block increase, decrease, or stay the same? Explain.

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QUESTION:

Problem 54P

A 0.505-kg block slides on a frictionless horizontal surface with a speed of 1.18 m/s. The block encounters an unstretched spring and compresses it 23.2 cm before coming to rest. (a) What is the force constant of this spring? (b) For what length of time is the block in contact with the spring before it comes to rest? (c) If the force constant of the spring is increased, docs the time required to stop the block increase, decrease, or stay the same? Explain.

ANSWER:

Step 1 of 3

Part a

We are required to calculate the force constant of the given spring.

Given:

The mass of the block $m=0.505$ kg

The speed of the block is $v=1.18$ m/s

The compression of the spring is $x=23.2$ cm = 0.232 m

The kinetic energy of the block is $KE=\frac{1}{2}m{v}^{2}$

$KE=\frac{1}{2}\times{}0.505\times{}(1.18{)}^{2}$ J

$KE=0.351$ J

If the spring has a force constant of $k$ , the spring potential energy is

$PE=\frac{1}{2}k{x}^{2}$

$PE=\frac{1}{2}k\times{}(0.232{)}^{2}$ J

Equating the two energy equations,

$0.351=\frac{1}{2}k\times{}(0.232{)}^{2}$

$k=\frac{2\times{}0.35}{(0.232{)}^{2}}$ N/m

$k=13.04$ N/m

The spring has a force constant of 13.04 N/m.

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