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Snowball A \(1.50 \mathrm{~kg}\) snowball is fired from a cliff \(12.5 \mathrm{~m}\) high with an initial velocity of \(14.0 \mathrm{~m} / \mathrm{s}\), directed \(41.0^{\circ}\) above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

Short Answer

Expert verified
(a) 183.75 J, (b) -183.75 J, (c) -183.75 J

Step by step solution

01

Title - Find the height of the fall

The snowball falls from a height of 12.5 m. To find the work done by the gravitational force, we first determine the total vertical distance covered by the snowball.
02

Title - Calculate work done by gravitational force

Work done by gravity is given by the formula: \[ W = mgh \]Where:m = mass of the snowball (1.50 kg)g = acceleration due to gravity (9.8 m/s²)h = height of the cliff (12.5 m)Substitute in the values:\[ W = 1.50 \times 9.8 \times 12.5 = 183.75 \text{ J} \]Thus, the work done by the gravitational force is 183.75 J.
03

Title - Determine the change in gravitational potential energy

The change in gravitational potential energy (\Delta U) is given by:\[ \Delta U = U_f - U_i \]Where:\(U_i\) is the initial potential energy and \(U_f\) is the final potential energy.Given \( U_f = 0\) (at ground level) and \( U_i = mgh = 183.75 J\), we get:\[ \Delta U = 0 - 183.75 = -183.75 \text{ J} \]Therefore, the change in gravitational potential energy is -183.75 J.
04

Title - Calculate potential energy at the ground

If the gravitational potential energy is taken to be zero at the height of the cliff, the value when the snowball reaches the ground is simply calculated to match the work done by gravity since potential energy decreases as height decreases.
05

Title - Conclusion

The gravitational potential energy at ground level, given it is zero at the height of the cliff, is -183.75 J. This confirms our earlier steps.

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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 the attractive force that pulls objects towards the center of the Earth. It acts on every object with mass and is given by Newton's law of gravitation. In this problem, the gravitational force acts on the snowball pulling it downward. The force of gravity can be calculated using the equation:

\( F = mg \)
where:
  • \( m \) is the mass of the object (in kg),
  • \( g \) is the acceleration due to gravity (9.8 m/s²).

For our snowball, the gravitational force pulling it downwards would be:

\( F = 1.50 \times 9.8 = 14.7 \text{ N} \)

This constant force is what causes the snowball to accelerate downwards and impacts the work done and energy calculations.
Potential Energy
Potential energy is the energy stored in an object due to its position. Gravitational potential energy depends on the height of an object relative to some reference point. The formula to calculate it is:

\( U = mgh \)

Here:
  • \( m \) is the mass of the object (1.50 kg),
  • \( g \) is the acceleration due to gravity (9.8 m/s²),
  • \( h \) is the height above the reference point (12.5 m in the case of the cliff).

The initial potential energy of the snowball at the top of the cliff can be calculated as:

\( U = 1.50 \times 9.8 \times 12.5 = 183.75 \text{ J} \)

This potential energy will change as the snowball falls.
Work Done by Gravity
Work done by gravity is the energy transferred to or from an object via the force of gravity as it moves. It can be calculated using the work-energy theorem, which states:

\( W = F \times d \times \text{cos(θ)} \)

In our scenario,
  • The force (F) is the weight of the snowball (14.7 N),
  • The displacement (d) is the vertical distance fallen (12.5 m),
  • \(\theta\) is 0 degrees since the force and displacement are in the same direction.

This simplifies the work done by gravity to:

\( W = mgh = 1.50 \times 9.8 \times 12.5 = 183.75 \text{ J} \)

This means gravity does 183.75 J of work on the snowball as it falls.
Kinematics
Kinematics studies the motion of objects without considering the forces that cause this motion. In this problem, kinematics elements such as initial velocity and angle play a role. The initial velocity (14.0 m/s) and the angle (41.0 degrees) of the snowball's launch affect how it moves. The snowball's trajectory can be split into horizontal and vertical components using trigonometry:
  • Vertical component: \( v_{y} = v \sin(\theta) = 14.0 \sin(41.0°) \)
  • Horizontal component: \( v_{x} = v \cos(\theta) = 14.0 \cos(41.0°) \)

These components are crucial to calculate further details like maximum height reached or time of flight, but for our energy calculations, they primarily provide context for how the initial conditions influence the snowball's motion.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In our scenario, the snowball's total mechanical energy (sum of kinetic and potential energy) remains constant if we neglect air resistance. Initially, the snowball has potential energy (due to its height) and kinetic energy (due to its velocity).

As it falls, potential energy is converted into kinetic energy. The conservation of energy equation can be written as:

\( U_i + K_i = U_f + K_f \)

where:
  • \( U_i \) is initial potential energy,
  • \( K_i \) is initial kinetic energy,
  • \( U_f \) is final potential energy,
  • \( K_f \) is final kinetic energy.

At ground level, the potential energy is zero if we take the reference height at ground level. The initial potential energy (183.75 J) has transformed into an equivalent amount of kinetic energy when the snowball hits the ground.

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Most popular questions from this chapter

Block Slides Down an Incline In Fig. \(10-48\), a block is moved down an incline a distance of \(5.0 \mathrm{~m}\) from point \(A\) to point \(B\) by a force \(\vec{F}\) that is parallel to the incline and has magnitude \(2.0 \mathrm{~N}\). The magnitude of the frictional force acting on the block is \(10 \mathrm{~N}\). If the kinetic energy of the block increases by \(35 \mathrm{~J}\) between \(A\) and \(B\), how much work is done on the block by the gravitational force as the block moves from \(A\) to \(B\) ?

Alpha Particle An alpha particle (mass \(4 \mathrm{u}\) ) experiences an elastic head-on collision with a gold nucleus (mass \(197 \mathrm{u}\) ) that is originally at rest. (The symbol u represents the atomic mass unit.) What percentage of its original kinetic energy does the alpha particle lose?

Mount Everest The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

Block on Incline Collides with Spring In Fig. \(10-37\), a \(12 \mathrm{~kg}\) block is released from rest on a \(30^{\circ}\) frictionless incline. Below the block is a spring that can be compressed \(2.0 \mathrm{~cm}\) by a force of \(270 \mathrm{~N}\). The block momentarily stops when it compresses the spring by \(5.5 \mathrm{~cm} .\) (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches spring?

Spring at the Top of an Incline a spring with spring constant \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a \(37.0^{\circ}\) frictionless incline. The lower end of the incline is \(1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

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