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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 body of mass \(m\) moves in one dimension under the influence of a force, \(F(x)\), which depends only on the body's position. a) Prove that Newton's Second Law and the law of conservation of energy for this body are exactly equivalent. b) Explain, then, why the law of conservation of energy is considered to be of greater significance than Newton's Second Law.

A spring with a spring constant of \(500 . \mathrm{N} / \mathrm{m}\) is used to propel a 0.500 -kg mass up an inclined plane. The spring is compressed \(30.0 \mathrm{~cm}\) from its equilibrium position and launches the mass from rest across a horizontal surface and onto the plane. The plane has a length of \(4.00 \mathrm{~m}\) and is inclined at \(30.0^{\circ} .\) Both the plane and the horizontal surface have a coefficient of kinetic friction with the mass of \(0.350 .\) When the spring is compressed, the mass is \(1.50 \mathrm{~m}\) from the bottom of the plane. a) What is the speed of the mass as it reaches the bottom of the plane? b) What is the speed of the mass as it reaches the top of the plane? c) What is the total work done by friction from the beginning to the end of the mass's motion?

A high jumper approaches the bar at \(9.0 \mathrm{~m} / \mathrm{s}\). What is the highest altitude the jumper can reach, if he does not use any additional push off the ground and is moving at \(7.0 \mathrm{~m} / \mathrm{s}\) as he goes over the bar?

A truck of mass 10,212 kg moving at a speed of \(61.2 \mathrm{mph}\) has lost its brakes. Fortunately, the driver finds a runaway lane, a gravel-covered incline that uses friction to stop a truck in such a situation; see the figure. In this case, the incline makes an angle of \(\theta=40.15^{\circ}\) with the horizontal, and the gravel has a coefficient of friction of 0.634 with the tires of the truck. How far along the incline \((\Delta x)\) does the truck travel before it stops?
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