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Chapter 6: Problem 45

A spring with \(k=10.0 \mathrm{~N} / \mathrm{cm}\) is initially stretched \(1.00 \mathrm{~cm}\) from its equilibrium length. a) How much more energy is needed to further stretch the spring to \(5.00 \mathrm{~cm}\) beyond its equilibrium length? b) From this new position, how much energy is needed to compress the spring to \(5.00 \mathrm{~cm}\) shorter than its equilibrium position?

Short Answer

Expert verified
Answer: The additional energy needed to stretch the spring to 5.00 cm beyond its equilibrium position is 120.0 Joules. The energy needed to compress the spring to 5.00 cm shorter than its equilibrium position from the new position is 375.0 Joules.

Step by step solution

01

Calculate initial potential energy

First, calculate the initial potential energy (PEi) stored in the spring when it is stretched 1.00 cm from its equilibrium position using the formula \(PE = \frac{1}{2}kx^2\). \(PEi = \frac{1}{2}(10.0 \mathrm{~N} / \mathrm{cm})(1.00 \mathrm{~cm})^2 = 5.00 \mathrm{J}\) The initial potential energy is 5.00 Joules.
02

Calculate final potential energy for part a

Next, calculate the final potential energy (PEa) stored in the spring when it is stretched to 5.00 cm beyond the equilibrium position using the formula \(PE = \frac{1}{2}kx^2\). \(PEa = \frac{1}{2}(10.0 \mathrm{~N} / \mathrm{cm})(5.00 \mathrm{~cm})^2 = 125.0 \mathrm{J}\) The final potential energy for part a is 125.0 Joules.
03

Find the additional energy needed for part a

To find the additional energy needed to stretch the spring, subtract the initial potential energy (PEi) from the final potential energy (PEa). Additional energy for part a = \(PEa - PEi = 125.0 \mathrm{J} - 5.00 \mathrm{J} = 120.0 \mathrm{J}\) The additional energy needed to stretch the spring to 5.00 cm beyond its equilibrium position is 120.0 Joules.
04

Calculate final potential energy for part b

Now, calculate the final potential energy (PEb) stored in the spring when it is compressed 5.00 cm shorter than its equilibrium position from the new position using the formula \(PE = \frac{1}{2}kx^2\). \(PEb = \frac{1}{2}(10.0 \mathrm{~N} / \mathrm{cm})(-10.00 \mathrm{~cm})^2 = 500.0 \mathrm{J}\) The final potential energy for part b is 500.0 Joules.
05

Find the energy needed for part b

To find the energy needed to compress the spring, subtract the final potential energy (PEa) from the final potential energy (PEb). Energy needed for part b = \(PEb - PEa = 500.0 \mathrm{J} - 125.0 \mathrm{J} = 375.0 \mathrm{J}\) The energy needed to compress the spring to 5.00 cm shorter than its equilibrium position from the new position is 375.0 Joules.

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Most popular questions from this chapter

A spring has a spring constant of \(80 \mathrm{~N} / \mathrm{m}\). How much potential energy does it store when stretched by \(1.0 \mathrm{~cm} ?\) a) \(4.0 \cdot 10^{-3}\) J b) \(0.40 \mathrm{~J}\) c) 80 d) \(800 \mathrm{~J}\) e) \(0.8 \mathrm{~J}\)

A classmate throws a \(1.0-\mathrm{kg}\) book from a height of \(1.0 \mathrm{~m}\) above the ground straight up into the air. The book reaches a maximum height of \(3.0 \mathrm{~m}\) above the ground and begins to fall back. Assume that \(1.0 \mathrm{~m}\) above the ground is the reference level for zero gravitational potential energy. Determine a) the gravitational potential energy of the book when it hits the ground. b) the velocity of the book just before hitting the ground.

Calculate the force \(F(y)\) associated with each of the following potential energies: a) \(U=a y^{3}-b y^{2}\) b) \(U=U_{0} \sin (c y)\)

a) If you are at the top of a toboggan run that is \(40.0 \mathrm{~m}\) high, how fast will you be going at the bottom, provided you can ignore friction between the sled and the track? b) Does the steepness of the run affect how fast you will be going at the bottom? c) If you do not ignore the small friction force, does the steepness of the track affect the value of the speed at the bottom?

A uniform chain of total mass \(m\) is laid out straight on a frictionless table and held stationary so that one-third of its length, \(L=1.00 \mathrm{~m},\) is hanging vertically over the edge of the table. The chain is then released. Determine the speed of the chain at the instant when only one-third of its length remains on the table.
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