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Chapter 6: Problem 37

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

Short Answer

Expert verified
Answer: The initial speed of the ball when it was thrown is approximately 9.95 m/s.

Step by step solution

01

Identifying the given information

We are given the maximum height reached by the ball, which is \(5.00 \mathrm{~m}\). Our task is to determine its initial speed when it was thrown.
02

Understanding the conservation of mechanical energy

The conservation of mechanical energy states that the total mechanical energy (potential + kinetic) in a closed system remains constant. In this case, we'll consider the ball's mechanical energy just when it is thrown and when it reaches its maximum height.
03

Applying the conservation of mechanical energy

When the ball is thrown, it has an initial kinetic energy and zero potential energy. At the maximum height, its potential energy is at the maximum, and its kinetic energy is zero since its velocity is zero. We can write the equation for the conservation of mechanical energy as: \(K_i + U_i = K_f + U_f\) where \(K_i\) - Initial kinetic energy \(U_i\) - Initial potential energy \(K_f\) - Final kinetic energy \(U_f\) - Final potential energy Since the initial potential energy is zero (let's consider the ground level as our reference), and the final kinetic energy is zero (at max height, the ball isn't moving), the equation becomes: \(K_i = U_f\)
04

Calculating the initial and final energies

The formula for the kinetic energy of an object is: \(K = \frac{1}{2}mv^2\) where \(m\) is the mass and \(v\) is the velocity. The formula for the gravitational potential energy is: \(U = mgh\) where \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)) and \(h\) is the height. Now, according to our energy conservation equation, we have: \(\frac{1}{2}mv^2 = mgh\)
05

Solving for the initial speed

We can cancel the mass \(m\) from both sides since it is a common factor: \(\frac{1}{2}v^2 = gh\) Now, we can solve for the initial speed (\(v\)) by isolating it: \(v^2 = 2gh\) \(v = \sqrt{2gh}\) Substitute the given height (\(h = 5.00\mathrm{~m}\)) and acceleration due to gravity (\(g = 9.81 \mathrm{~m/s^2}\)): \(v = \sqrt{(2)(9.81 \mathrm{~m/s^2})(5.00\mathrm{~m})}\) \(v \approx 9.95 \mathrm{~m/s}\) So, the ball's initial speed when it was thrown is approximately \(9.95 \mathrm{~m/s}\).

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Most popular questions from this chapter

Can a unique potential energy function be identified with a particular conservative force?

A block of mass \(5.0 \mathrm{~kg}\) slides without friction at a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) on a horizontal table surface until it strikes and sticks to a mass of \(4.0 \mathrm{~kg}\) attached to a horizontal spring (with spring constant of \(k=2000.0 \mathrm{~N} / \mathrm{m}\) ), which in turn is attached to a wall. How far is the spring compressed before the masses come to rest? a) \(0.40 \mathrm{~m}\) b) \(0.54 \mathrm{~m}\) c) \(0.30 \mathrm{~m}\) d) \(0.020 \mathrm{~m}\) e) \(0.67 \mathrm{~m}\)

A 0.500 -kg mass is attached to a horizontal spring with \(k=100 . \mathrm{N} / \mathrm{m}\). The mass slides across a frictionless surface. The spring is stretched \(25.0 \mathrm{~cm}\) from equilibrium, and then the mass is released from rest. a) Find the mechanical energy of the system. b) Find the speed of the mass when it has moved \(5.00 \mathrm{~cm}\). c) Find the maximum speed of the mass.

A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1600 . \mathrm{N} / \mathrm{m} .\) The clay compresses the spring a certain maximum amount before momentarily stopping a) Find the maximum compression of the spring. b) Find the total work done on the clay during the spring's compression.

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)?
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