/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Someone drops a \(50-\mathrm{g}\... [FREE SOLUTION] | 91Ó°ÊÓ

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Someone drops a \(50-\mathrm{g}\) pebble off of a docked cruise ship, \(70.0 \mathrm{m}\) from the water line. A person on a dock \(3.0 \mathrm{m}\) from the water line holds out a net to catch the pebble. (a) How much work is done on the pebble by gravity during the drop? (b) What is the change in the gravitational potential energy during the drop? If the gravitational potential energy is zero at the water line, what is the gravitational potential energy (c) when the pebble is dropped? (d) When it reaches the net? What if the gravitational potential energy was 30.0 Joules at water level? (e) Find the answers to the same questions in (c) and (d).

Short Answer

Expert verified
(a) The work done by gravity on the pebble during the drop is \(34.285\,\mathrm{J}\). (b) The change in gravitational potential energy during the drop is also \(34.285\,\mathrm{J}\). (c) The gravitational potential energy when the pebble is dropped is \(34.285\,\mathrm{J}\). (d) The gravitational potential energy when the pebble reaches the net is \(1.4705\,\mathrm{J}\). (e) With the given initial potential energy of \(30.0\,\mathrm{J}\), the gravitational potential energy when the pebble is dropped is \(64.285\,\mathrm{J}\), and when it reaches the net, it is \(31.4705\,\mathrm{J}\).

Step by step solution

01

Find the work done by gravity on the pebble

To find the work done by gravity on the pebble during the drop, we can use the formula: \( W = mgh \) where \(m = 50\,\mathrm{g} = 0.05\,\mathrm{kg}\) (Convert the mass from grams to kg), \(g = 9.81\,\mathrm{m/s^2}\) (acceleration due to gravity), \(h = 70\,\mathrm{m}\) (height above the waterline). Substituting the values, we have: \( W = (0.05\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(70\,\mathrm{m})\) \( W = 34.285\,\mathrm{J}\) The work done by gravity on the pebble during the drop is \(34.285\,\mathrm{J}\).
02

Find the change in the gravitational potential energy

To find the change in gravitational potential energy during the drop, we will use the same formula as in Step 1: \( \Delta U = mgh \) Since the formula is the same, the change in gravitational potential energy during the drop is also \(34.285\,\mathrm{J}\).
03

Find the gravitational potential energy when the pebble is dropped

Since in this part it's mentioned that the gravitational potential energy is zero at the waterline, that means when the pebble is dropped, the gravitational potential energy is: \( U_1 = mgh \) As we have calculated previously, the gravitational potential energy when the pebble is dropped is \(34.285\,\mathrm{J}\).
04

Find the gravitational potential energy when it reaches the net

To find the gravitational potential energy when the pebble reaches the net, we need to use the height from the waterline to the person on the dock holding the net, which is \(3\,\mathrm{m}\). This time, the gravitational potential energy becomes: \( U_2 = mg(3\,\mathrm{m})\) Substituting the values, we have: \( U_2 = (0.05\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(3\,\mathrm{m}) \) \( U_2 = 1.4705\,\mathrm{J}\) The gravitational potential energy when the pebble reaches the net is \(1.4705\,\mathrm{J}\).
05

Calculate gravitational potential energy with the given initial potential energy

The initial potential energy is given as \(30.0\,\mathrm{J}\) at water level. Therefore, we will add this value to the potential energies calculated in Steps 3 and 4. For the gravitational potential energy when the pebble is dropped, we have: \( U'_1 = U_1 + 30.0\,\mathrm{J} = 34.285\,\mathrm{J} + 30.0\,\mathrm{J} = 64.285\,\mathrm{J}\) For the gravitational potential energy when the pebble reaches the net, we have: \( U'_2 = U_2 + 30.0\,\mathrm{J} = 1.4705\,\mathrm{J} + 30.0\,\mathrm{J} = 31.4705\,\mathrm{J}\) With the given initial potential energy of \(30.0\,\mathrm{J}\), the gravitational potential energy when the pebble is dropped is \(64.285\,\mathrm{J}\), and when it reaches the net, it is \(31.4705\,\mathrm{J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. It is dependent on three main factors: the object's mass (\( m \)), the height (\( h \)) of the object above a reference level, and the acceleration due to gravity (\( g \)), which is approximately \( 9.81 \,\mathrm{m/s^2} \) on Earth. This potential energy is mathematical expressed as \( U = mgh \).

For instance, in the problem we are exploring, when the pebble is held \( 70 \) meters above the waterline, its gravitational potential energy can be calculated. When calculating or comparing gravitational potential energy, it's crucial to set a baseline, or point of zero potential energy. In our example, the waterline is initially set as this zero point.

Understanding these concepts is essential, not only for calculating the gravitational potential energy at various positions but also for comprehending how energy transformations occur during a free fall.
Work Done by Gravity
Work done by gravity involves the energy transferred to or from an object by the gravitational force as it moves a certain distance. It is closely related to gravitational potential energy because any change in gravitational potential energy corresponds to work being done by gravity. The formula to calculate this work is similar to that for gravitational potential energy: \( W = mgh \). This signifies that the work done by gravity is essentially the energy shift as an object moves from one point to another vertically.

In our exercise, the pebble falls from a height of \( 70 \) meters to a net located \( 3 \) meters above the water. By calculating \( mgh \) for this height difference, we find that the work done by gravity is \( 34.285 \, \mathrm{J} \). This represents the energy gravity has transferred to the pebble during its descent, emphasizing the powerful role gravity plays in influencing motion and energy distribution.
Physics Problems
Solving physics problems, like the one involving the falling pebble, requires understanding the underlying principles of physics concepts such as gravitational potential energy and work.

A typical problem-solving approach includes identifying:
  • All given data, like the pebble's mass and height.
  • The formulas applicable to the problem, such as \( W = mgh \).
  • The conditions, like what is defined as zero potential energy.
Developing a systematic approach to physics problems helps in breaking down complex concepts into manageable calculations, ensuring a step-by-step resolution. Most importantly, interpreting the physical meaning of the results is key, as seen when calculating how energy changes and is impacted by various positions and influences. Practicing these components strengthens proficiency not just in dealing with numbers but in grasping how fundamental forces like gravity practically operate in our world.

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Most popular questions from this chapter

Consider a block of mass 0.200 kg attached to a spring of spring constant \(100 \mathrm{N} / \mathrm{m}\). The block is placed on a frictionless table, and the other end of the spring is attached to the wall so that the spring is level with the table. The block is then pushed in so that the spring is compressed by \(10.0 \mathrm{cm} .\) Find the speed of the block as it crosses (a) the point when the spring is not stretched, (b) \(5.00 \mathrm{cm}\) to the left of point in (a), and (c) \(5.00 \mathrm{cm}\) to the right of point in (a).

A particle of mass \(2.0 \mathrm{kg}\) moves under the influence If its speed at of the force \(F(x)=\left(-5 x^{2}+7 x\right)\) N. \(x=-4.0 \mathrm{m} \quad\) is \(\quad v=20.0 \mathrm{m} / \mathrm{s}, \quad\) what is its speed at \(x=4.0 \mathrm{m} ?\)

A projectile of mass 2 kg is fired with a speed of 20 \(\mathrm{m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with respect to the horizontal. (a) Calculate the initial total energy of the projectile given that the reference point of zero gravitational potential energy at the launch position. (b) Calculate the kinetic energy at the highest vertical position of the projectile. (c) Calculate the gravitational potential energy at the highest vertical position. (d) Calculate the maximum height that the projectile reaches. Compare this result by solving the same problem using your knowledge of projectile motion.

A cat's crinkle ball toy of mass \(15 \mathrm{g}\) is thrown straight up with an initial speed of \(3 \mathrm{m} / \mathrm{s}\). Assume in this problem that air drag is negligible. (a) What is the kinetic energy of the ball as it leaves the hand? (b) How much work is done by the gravitational force during the ball's rise to its peak? (c) What is the change in the gravitational potential energy of the ball during the rise to its peak? (d) If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when it reaches the maximum height? (e) What if the gravitational potential energy is taken to be zero at the maximum height the ball reaches, what would the gravitational potential energy be when it leaves the hand? (f) What is the maximum height the ball reaches?

In a coyote/Road Runner cartoon clip (https://openstaxcollege.org/I/21coyroadcarcl) , a spring expands quickly and sends the coyote into a rock. If the spring extended \(5 \mathrm{m}\) and sent the coyote of mass 20 kg to a speed of \(15 \mathrm{m} / \mathrm{s},\) (a) what is the spring constant of this spring? (b) If the coyote were sent vertically into the air with the energy given to him by the spring, how high could he go if there were no non-conservative forces?

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