/*! 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 16 You throw a \(20-\mathrm{N}\) ro... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 16

You throw a \(20-\mathrm{N}\) rock vertically into the air from ground level. You observe that when it is 15.0 \(\mathrm{m}\) above the ground, it is traveling at 25.0 \(\mathrm{m} / \mathrm{s}\) upward. Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

Short Answer

Expert verified
(a) Initial speed is 28.6 m/s. (b) Maximum height is 50 m.

Step by step solution

01

Identify the Known Quantities

First, identify all the known quantities. The rock has a weight (force due to gravity) of \(20 \text{ N}\), meaning its mass \(m = \frac{20}{9.8} \approx 2.04 \text{ kg}\). At 15.0 \text{ m}\, the rock's velocity is \(25.0 \text{ m/s}\) upward. Initial kinetic energy is unknown and needs to be found, as well as the maximum height it reaches.
02

Determine Initial Speed Using Work-Energy Theorem

Apply the work-energy theorem: the change in kinetic energy is equal to the work done by the gravitational force. Calculate the kinetic energy at \(15.0 \text{ m}\) using \(K = \frac{1}{2}mv^2\). The work done by gravity from the ground to \(15.0 \text{ m}\) is \( -mgh \), where \( h = 15 \text{ m}\). Set the initial kinetic energy plus work done equal to the kinetic energy at \(15.0 \text{ m}\). Solve for initial velocity \(v_i\).\[ \frac{1}{2}mv_i^2 - mgh = \frac{1}{2}mv^2 \]
03

Calculate Maximum Height

To find the maximum height, use energy conservation: Total mechanical energy when rock leaves the ground is equal to the total mechanical energy at max height (potential energy, since velocity is zero there). Set initial kinetic energy plus gravitational potential energy at ground level equal to gravitational potential energy at max height \(h_{max}\).\[ \frac{1}{2}mv_i^2 = mgh_{max} \]
04

Substitute and Solve Equations

For initial speed, substitute known values into the equations and solve for \(v_i\): \[ \frac{1}{2}\times 2.04 \times v_i^2 - 2.04 \times 9.8 \times 15 = \frac{1}{2}\times 2.04 \times (25)^2 \]Solve for \(v_i\). Then, using the speed from this solution, calculate the maximum height:\[h_{max} = \frac{\frac{1}{2}mv_i^2}{mg}\]

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Key Concepts

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

Kinetic Energy
Kinetic energy is an essential player in the work-energy theorem. It represents the energy a body possesses due to its motion. The faster an object moves, the higher its kinetic energy. Kinetic energy is given by the formula:
  • The equation: \( K = \frac{1}{2}mv^2 \)
  • Where \( m \) is the mass and \( v \) is the velocity
In the context of the problem, we're looking at the kinetic energy that the rock possesses at different points in its journey. Initially, on the ground, its kinetic energy helps conquer gravity to elevate the rock. As it climbs, this energy changes, influenced by the work done against gravity (the opposing force acting upon it). By understanding how kinetic energy shifts and is related to work done, we can determine the rock's speed when it was just launched.
Potential Energy
Potential energy is the stored energy that an object has due to its position or height. For objects near Earth鈥檚 surface, gravitational potential energy is most relevant. It depends on three factors:
  • Weight of the object \( mg \)
  • Height above a reference point \( h \)
  • The gravitational constant \( g \), approximately \(9.8 \, \text{m/s}^2\)
The formula for gravitational potential energy is:\[ U = mgh \]In the given problem, the rock possesses potential energy as it rises 15 m above the ground. This energy increases as it climbs until it reaches its peak. At maximum height, all the kinetic energy initially provided has transformed into potential energy. Understanding this conversion helps solve for how high the rock traveled following its lift-off from the ground.
Conservation of Energy
The principle of energy conservation states that the total mechanical energy of a system remains constant provided no external forces are doing work. This involves the transformation between kinetic and potential energy. In simpler terms, energy is neither created nor destroyed, merely converted from one form to another. In our scenario, the rock's initial kinetic energy when launched becomes partly converted into gravitational potential energy as it reaches its highest point above the ground. At maximum height, all kinetic energy has transformed into potential energy, causing the rock to momentarily stop before plummeting back down. Employing the conservation of energy principle allows us to solve for both the initial speed when leaving the ground and the maximum height of the rock.
Physics Problem Solving
Solving physics problems systematically requires a clear method. Here鈥檚 how you can approach problems like the one about the thrown rock using the work-energy theorem:
  • Identify known quantities and relevant formulas.
  • Apply the work-energy theorem: calculate the change in kinetic energy and equate it to work done.
  • Use energy conservation to relate initial conditions with final conditions, such as speed or height.
  • Substitute known values and solve the equations step-by-step.
This method works well for problems involving kinetic and potential energy changes. By breaking down each step, physics becomes more approachable and easier to solve.

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Most popular questions from this chapter

Stopping Distance. A car is traveling on a level road with speed \(v_{0}\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of \(v_{0}, g,\) and the coefficient of kinetic friction \(\mu_{x}\) between the tires and the road. (b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

Chin-Ups. While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m}\) . (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-k g\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be 10 \(\mathrm{g}\) , and assume the wings move an average downward distance of 1.0 \(\mathrm{cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass 70 \(\mathrm{g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A \(70-\mathrm{kg}\) athlete can maintain a power output of 1.4 \(\mathrm{kW}\) for no more than a few seconds; the steady power output of a typical athlete is only 500 \(\mathrm{W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

A block of ice with mass 6.00 \(\mathrm{kg}\) is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force \(\vec{F}\) to it. As a result, the block moves along the \(x\) -axis such that its position as a function of time is given by \(x(t)=\alpha t^{2}+\beta t^{3}\) , where \(\alpha=0.200 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=0.0200 \mathrm{m} / \mathrm{s}^{3}\) . (a) Calculate the velocity of the object when \(t=4.00 \mathrm{s}\) . (b) Calculate the magnitude of \(\overrightarrow{\boldsymbol{F}}\) when \(t=4.00 \mathrm{s}\) . (c) Calculate the work done by the force \(\overrightarrow{\boldsymbol{F}}\) during the first 4.00 \(\mathrm{s}\) of the motion.
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