/*! 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 26 A shot-putter puts a shot (weigh... [FREE SOLUTION] | 91Ó°ÊÓ

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A shot-putter puts a shot (weight \(=71.1 \mathrm{~N}\) ) that leaves his hand at a distance of \(1.52 \mathrm{~m}\) above the ground. (a) Find the work done by the gravitational force when the shot has risen to a height of \(2.13 \mathrm{~m}\) above the ground. Include the correct sign for the work. (b) Determine the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy of the shot

Short Answer

Expert verified
The work done by gravity is -43.371 J; the change in potential energy is 43.371 J.

Step by step solution

01

Understand the Problem

We need to calculate the work done by gravity on a shot put as it moves from an initial height of \(1.52 \text{ m}\) to a final height of \(2.13 \text{ m}\). We also need to determine the change in gravitational potential energy (\(\Delta \text{PE}\)). The weight \(W\) of the shot put is the force due to gravity \(71.1 \text{ N}\).
02

Calculate Work Done by Gravitational Force

The work \(W_{g}\) done by the gravitational force is given by: \[ W_{g} = -F_g \cdot \Delta h \] where \(F_g = 71.1 \text{ N}\) is the gravitational force, and \(\Delta h = 2.13 \text{ m} - 1.52 \text{ m}\) is the change in height. Substituting the values, \[ \Delta h = 0.61 \text{ m} \] \[ W_{g} = -(71.1 \times 0.61) \] Calculate the value to find \(W_{g}\).
03

Work Calculation

After calculating, we find: \[ W_{g} = -43.371 \text{ J} \] The negative sign indicates that the work is done against the direction of gravity (since the object is moving upward).
04

Determine Change in Potential Energy

The change in gravitational potential energy \(\Delta \text{PE}\) is defined by: \[ \Delta \text{PE} = mgh_f - mgh_0 \] Here, \(h_f = 2.13 \text{ m}\) and \(h_0 = 1.52 \text{ m}\). Using \(W = mg\), where \(mg = 71.1 \text{ N}\), we have: \[ \Delta \text{PE} = W \times \Delta h = 71.1 \times 0.61 \] Solve for the change in potential energy.
05

Calculate Change in Potential Energy

Calculate the change: \[ \Delta \text{PE} = 43.371 \text{ J} \] This result confirms that the potential energy increases by this amount as the object is lifted.

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Key Concepts

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

Gravitational Force
Gravitational force is one of the fundamental forces of nature that attracts two objects toward each other. In the context of our exercise, it's the force acting downward due to the weight of the shot put, essentially pulling it towards the Earth. The gravitational force can be calculated by using the formula: \( F_g = m imes g \), where \( m \) is mass and \( g \) is the acceleration due to gravity, typically \( 9.81 \, \text{m/s}^2 \).
Since the weight of the shot put is given directly as \( 71.1 \, \text{N} \), this is already the gravitational force acting on it. The gravitational force always acts in a downward direction, influencing objects to fall toward the Earth. This opposes any upward movement, resulting in work done against gravity when the object is lifted against this force.
  • The gravitational force depends on both the mass of an object and the gravitational acceleration of Earth.
  • When lifting an object, you are working against this gravitational pull.
  • This force makes it possible to calculate changes in energy, as seen when determining work done and potential energy change.
Gravitational Potential Energy
Gravitational potential energy (GPE) is energy that an object possesses because of its position in a gravitational field. In simpler terms, it is the energy stored due to an object's height above the ground. This stored energy can be calculated using the formula: \( \ GPE = mgh \ \) where \( m \) is the mass of the object, \( g \) is the gravitational acceleration, and \( h \) is the height of the object.
In our problem, we calculate the change in gravitational potential energy as the shot put ascends from \( 1.52 \, \text{m} \) to \( 2.13 \, \text{m} \). The change in potential energy corresponds to the difference between the final and initial energy states, given by \( \ \Delta \text{PE} = mgh_f - mgh_0 \ \), which simplifies to \( \ \Delta \text{PE} = mg \times (h_f - h_0) \ \).
  • GPE is proportional to height; the higher the object, the more GPE it has.
  • An increase in height leads to an increase in GPE, reflecting the stored energy due to gravitational work.
  • The distinct advantage of GPE is that it offers insights into the energy dynamics involved in lifting or dropping something.
Physics Problem Solving
Solving physics problems systematically requires understanding of both theoretical concepts and practical calculations. In this particular exercise, the concepts of gravitational force and gravitational potential energy are applied in sequence to find numerical answers.
To solve such problems effectively, it's crucial to break them down into smaller, manageable steps:
  • Begin by fully understanding the problem – identify what is known and what needs to be found.
  • Label the given values and formulas that relate them, such as those for work (\( W = F \, \Delta h \)) and potential energy (\( \Delta \text{PE} = mgh \)).
  • Perform calculations systematically, checking each step for accuracy to ensure your final result is correct.
In our exercise, the task was to find both the work done by gravitational force and the change in gravitational potential energy as a shot put was lifted. Knowing how to apply the appropriate formulas at each step is critical.
This involves consistent practice and familiarity with the underlying physics principles. By following a structured approach, students can develop strong problem-solving skills that lay the groundwork for more advanced physics concepts.

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Most popular questions from this chapter

A \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=524 \mathrm{~N}\). A force \(\overrightarrow{\mathrm{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathrm{W}}\) and the normal force \(\overrightarrow{\mathrm{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is \(290 \mathrm{~m}\). What should be the magnitude of \(\overrightarrow{\mathbf{F}}\), so that the net work done by all the forces acting on the car is \(+150 \mathrm{~kJ}\) ?

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water \(5.00 \mathrm{~m}\) from the end of the slide in a time of \(0.500 \mathrm{~s}\) after leaving the slide. Ignoring friction and air resistance, find the height \(H\) in the drawing.

A basketball of mass \(0.60 \mathrm{~kg}\) is dropped from rest from a height of \(1.05 \mathrm{~m}\). It rebounds to a height of \(0.57 \mathrm{~m}\) (a) How much mechanical energy was lost during the collision with the floor? (b) A basketball player dribbles the ball from a height of \(1.05 \mathrm{~m}\) by exerting a constant downward force on it for a distance of \(0.080 \mathrm{~m}\). In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.05 \(\mathrm{m},\) what is the magnitude of the force?

A roller coaster \((375 \mathrm{~kg})\) moves from \(A(5.00 \mathrm{~m}\) above the ground) to \(B(20.0 \mathrm{~m}\) above the ground). Two nonconservative forces are present: friction does \(-2.00 \times 10^{4} \mathrm{~J}\) of work on the car, and a chain mechanism does \(+3.00 \times 10^{4} \mathrm{~J}\) of work to help the car up a long climb. What is the change in the car's kinetic energy, \(\Delta \mathrm{KE}=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{0}\), from \(A\) to \(B\) ?

One of the new events in the 2002 Winter Olympics was the sport of skeleton (see the photo). Starting at the top of a steep, icy track, a rider jumps onto a sled (known as a skeleton) and proceeds-belly down and head first-to slide down the track. The track has fifteen turns and drops \(104 \mathrm{~m}\) in elevation from top to bottom. (a) In the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed of the rider at the beginning of the run is relatively small and can be ignored. (b) In reality, the best riders reach the bottom with a speed of \(35.8 \mathrm{~m} / \mathrm{s}\) (about \(80 \mathrm{mi} / \mathrm{h}\) ). How much work is done on an \(86.0-\mathrm{kg}\) rider and skeleton by nonconservative forces?

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