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Chapter 6: Problem 80

A 0.100 -kg ball is dropped from a height of \(1.00 \mathrm{~m}\) and lands on a light (approximately massless) cup mounted on top of a light, vertical spring initially at its equilibrium position. The maximum compression of the spring is to be \(10.0 \mathrm{~cm}\). a) What is the required spring constant of the spring? b) Suppose you ignore the change in the gravitational energy of the ball during the 10 -cm compression. What is the percentage difference between the calculated spring constant for this case and the answer obtained in part (a)?

Short Answer

Expert verified
Answer: The spring constant of the light, vertical spring is 196.2 N/m. The percentage difference between the calculated spring constants considering and ignoring the change in gravitational energy during the 10-cm compression is approximately 0.10%.

Step by step solution

01

Noting the given information

We have the following information given in the problem: 1. Mass of the ball (m) = 0.100 kg 2. Height from which ball is dropped (h) = 1.00 m 3. Maximum compression of the spring (x) = 10.0 cm = 0.1 m (converted to meters for consistency)
02

Determine the initial and final energies

At the initial moment before dropping the ball, it only has gravitational potential energy: Initial total energy (E_initial) = Gravitational potential energy = m * g * h At the final moment when the spring is maximally compressed, the ball has both gravitational potential energy and spring potential energy: Final total energy (E_final) = Gravitational potential energy + Spring potential energy = (m * g * (h - x)) + (1/2 * k * x^2) Here k is the spring constant, which we need to find.
03

Apply conservation of mechanical energy

Since the mechanical energy is conserved, we can equate the initial and final energies: E_initial = E_final m * g * h = (m * g * (h - x)) + (1/2 * k * x^2)
04

Solve for the spring constant, k

We will now isolate k from the equation above and solve for it. m * g * h - m * g * (h - x) = (1/2 * k * x^2) k = (2 * (m * g * h - m * g * (h - x))) / x^2 Now substituting the given values to calculate k: k = (2 * (0.100 kg * 9.81 m/s^2 * 1.00 m - 0.100 kg * 9.81 m/s^2 * (1.00 m - 0.1 m))) / (0.1 m)^2 k = 196.2 N/m So, the required spring constant is k = 196.2 N/m.
05

Calculate the spring constant without considering the change in gravitational potential energy

Now, if we ignore the change in gravitational potential energy when the spring is compressed, we have to only consider the spring potential energy at the final moment: E_final' = Spring potential energy = (1/2 * k * x^2) E_initial = E_final' m * g * h = (1/2 * k * x^2) k' = (2 * (m * g * h)) / x^2 Now substituting the given values to calculate k': k' = (2 * (0.100 kg * 9.81 m/s^2 * 1.00 m)) / (0.1 m)^2 k' = 196.0 N/m So, the calculated spring constant without considering the change in gravitational potential energy is k' = 196.0 N/m.
06

Calculate the percentage difference

The percentage difference between the calculated spring constants k and k' can be calculated as follows: Percentage Difference = 100 * (|k - k'|) / k Percentage Difference = 100 * (|196.2 N/m - 196.0 N/m|) / 196.2 N/m Percentage Difference ≈ 0.10% The percentage difference between the calculated spring constants is approximately 0.10%.

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Most popular questions from this chapter

A ball of mass \(1.84 \mathrm{~kg}\) is dropped from a height \(y_{1}=$$1.49 \mathrm{~m}\) and then bounces back up to a height of \(y_{2}=0.87 \mathrm{~m}\) How much mechanical energy is lost in the bounce? The effect of air resistance has been experimentally found to be negligible in this case, and you can ignore it.

A roller coaster is moving at \(2.00 \mathrm{~m} / \mathrm{s}\) at the top of the first hill \((h=40.0 \mathrm{~m}) .\) Ignoring friction and air resistance, how fast will the roller coaster be moving at the top of a subsequent hill, which is \(15.0 \mathrm{~m}\) high?

What is the gravitational potential energy of a \(2.0-\mathrm{kg}\) book \(1.5 \mathrm{~m}\) above the floor?

A ball is thrown up in the air, reaching a height of \(5.00 \mathrm{~m}\). Using energy conservation considerations, determine its initial speed.

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.
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