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Chapter 6: Problem 28

a) If the gravitational potential energy of a 40.0 -kg rock is 500 . J relative to a value of zero on the ground, how high is the rock above the ground? b) If the rock were lifted to twice its original height, how would the value of its gravitational potential energy change?

Short Answer

Expert verified
Answer: The rock is placed at a height of approximately 1.27 meters above the ground. If the rock is lifted to twice its original height, its gravitational potential energy would double to 1000 J.

Step by step solution

01

Identify the Given Information and Formula

We are given the mass of the rock (m = 40 kg) and its gravitational potential energy (U = 500 J). The gravitational potential energy formula is: U = mgh where U is gravitational potential energy, m is mass of the object, g is the acceleration due to gravity (approximately 9.81 m/s^2), and h is the height of the object above the ground.
02

Solve for Height (h)

Using the formula U = mgh, we can solve for height (h). Rearrange the formula to isolate h: h = U / (mg) Now, plug in the given values to find h: h = (500 J) / (40 kg × 9.81 m/s^2) Calculate the value of h: h ≈ 1.27 m For part a), the rock is approximately 1.27 meters above the ground.
03

Calculate New Potential Energy

For part b), we have to find the new gravitational potential energy if the rock is lifted to twice its original height. If the original height is h, the new height will be 2h.
04

Substitute New Height Back into Formula

Use the formula for gravitational potential energy with the new height (2h): U_new = mg(2h)
05

Calculate New Potential Energy

Since we know the original height (h = 1.27 m), we can calculate the new potential energy at the doubled height: U_new = (40 kg)(9.81 m/s^2)(2 × 1.27 m) Calculate the value of U_new: U_new ≈ 1000 J For part b), the gravitational potential energy of the rock would double to 1000 J if it was lifted to twice its original height.

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Most popular questions from this chapter

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 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.

How much mechanical energy is lost to friction if a 55.0-kg skier slides down a ski slope at constant speed of \(14.4 \mathrm{~m} / \mathrm{s}\) ? The slope is \(123.5 \mathrm{~m}\) long and makes an angle of \(14.7^{\circ}\) with respect to the horizontal.

A large air-filled 0.100 -kg plastic ball is thrown up into the air with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\). At a height of \(3.00 \mathrm{~m}\) the ball's speed is \(3.00 \mathrm{~m} / \mathrm{s}\). What fraction of its original energy has been lost to air friction?

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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