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Chapter 10: Problem 46

. Frisbee A \(75 \mathrm{~g}\) Frisbee is thrown from a point \(1.1 \mathrm{~m}\) above the ground with a speed of \(12 \mathrm{~m} / \mathrm{s}\). When it has reached a height of \(2.1 \mathrm{~m}\), its speed is \(10.5 \mathrm{~m} / \mathrm{s}\). What was the reduction in the mechanical energy of the Frisbee- Earth system because of air drag?

Short Answer

Expert verified
The reduction in mechanical energy is 0.5306 J.

Step by step solution

01

- Identify given variables

Determine and list the given variables: - Mass of the frisbee, \(m = 75 \text{ g} = 0.075 \text{ kg}\)- Initial height from the ground, \(h_1 = 1.1 \text{ m}\)- Initial speed, \(v_1 = 12 \text{ m/s}\)- Final height, \(h_2 = 2.1 \text{ m}\)- Final speed, \(v_2 = 10.5 \text{ m/s}\)
02

- Calculate initial mechanical energy

Initial mechanical energy consists of kinetic energy (K.E) and potential energy (P.E).\[\text{Initial K.E} = \frac{1}{2} m v_1^2 = \frac{1}{2} \times 0.075 \text{ kg} \times (12 \text{ m/s})^2 = 5.4 \text{ J}\]\[\text{Initial P.E} = m g h_1 = 0.075 \text{ kg} \times 9.8 \text{ m/s}^2 \times 1.1 \text{ m} = 0.8085 \text{ J}\]\[\text{Total Initial Mechanical Energy} = 5.4 \text{ J} + 0.8085 \text{ J} = 6.2085 \text{ J}\]
03

- Calculate final mechanical energy

Final mechanical energy consists of kinetic energy (K.E) and potential energy (P.E).\[\text{Final K.E} = \frac{1}{2} m v_2^2 = \frac{1}{2} \times 0.075 \text{ kg} \times (10.5 \text{ m/s})^2 = 4.1344 \text{ J}\]\[\text{Final P.E} = m g h_2 = 0.075 \text{ kg} \times 9.8 \text{ m/s}^2 \times 2.1 \text{ m} = 1.5435 \text{ J}\]\[\text{Total Final Mechanical Energy} = 4.1344 \text{ J} + 1.5435 \text{ J} = 5.6779 \text{ J}\]
04

- Determine the reduction in mechanical energy

The reduction in mechanical energy is the difference between the initial and final mechanical energy.\[\text{Reduction in Mechanical Energy} = \text{Total Initial Mechanical Energy} - \text{Total Final Mechanical Energy}\]\[= 6.2085 \text{ J} - 5.6779 \text{ J} = 0.5306 \text{ J}\]

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Key Concepts

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

Kinetic Energy
Kinetic energy, often abbreviated as K.E, is the energy that an object possesses due to its motion. Whenever an object moves, it has kinetic energy. This energy can be calculated using the formula: \[ \text{K.E} = \frac{1}{2} m v^2 \] where \(m\) is the mass of the object and \(v\) is its velocity. In the exercise, the Frisbee鈥檚 initial kinetic energy was determined at the starting speed of 12 m/s and its mass of 0.075 kg, resulting in 5.4 J (Joules). As the Frisbee slows down to 10.5 m/s, its kinetic energy decreases to 4.1344 J. The difference indicates a loss in the system's kinetic energy as it moves upward.
Potential Energy
Potential energy, abbreviated as P.E, is the energy stored in an object due to its position relative to a particular reference point, often in a gravitational field. The formula to calculate gravitational potential energy is: \[ \text{P.E} = mgh \] where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately 9.8 m/s虏 on Earth's surface), and \(h\) is the height above the reference point. For the Frisbee, its initial potential energy at a height of 1.1 m was 0.8085 J. When it rises to 2.1 m, its potential energy increases to 1.5435 J. This rise reflects the energy gained as its height increases.
Air Drag
Air drag, also known as air resistance, is a force that opposes the motion of an object through the air. It depends on several factors, including the object's speed, surface area, and shape, as well as the air density. In our exercise, the Frisbee's mechanical energy decreases by 0.5306 J as it rises and slows down. This loss is attributed to air drag, which dissipated part of the system鈥檚 mechanical energy. Air drag converts mechanical energy into other forms, typically thermal energy, leading to a reduction in the total mechanical energy of the Frisbee-Earth system.

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Most popular questions from this chapter

You drop a \(2.00\) kg textbook to a friend who stands on the ground \(10.0 \mathrm{~m}\) below the textbook with outstretched hands \(1.50 \mathrm{~m}\) above the ground (Fig. \(10-26\) ). (a) How much work \(W^{\text {grav }}\) is done on the textbook by the gravitational force as it drops to your friend's hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the textbook-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) when the textbook (c) is released and (d) reaches the hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W\) grav (f) \(\Delta U,(\mathrm{~g}) U\) at the release point, and (h) \(U\) at the hands.

Two Bodies Collide Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B\), collide. The velocities before the collision are \(\vec{v}_{A 1}=(15 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(30 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and \(\vec{v}_{B 1}=(-10 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). After the collision, \(\vec{v}_{A 2}=\) \((-5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). (a) What is the final velocity of \(B ?\) (b) How much kinetic energy is gained or lost in the collision?

Elastic Collision A body of mass \(2.0 \mathrm{~kg}\) makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the \(2.0 \mathrm{~kg}\) body was \(4.0 \mathrm{~m} / \mathrm{s}\) ?

Block Drawn by Rope A \(3.57 \mathrm{~kg}\) block is drawn at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor by a rope. The force on the block from the rope has a magnitude of \(7.68 \mathrm{~N}\) and is directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

Block Dropped on a Spring Two A \(2.0 \mathrm{~kg}\) block is dropped from a height of \(40 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. 10-39). Find the maximum distance the spring is compressed.
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