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Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of \(6.6 \mathrm{m} / \mathrm{s} .\) Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Short Answer

Expert verified
(a) The hill is approximately 2.22 meters high. (b) The frozen juice can also reaches 6.6 m/s.

Step by step solution

01

Identify Known Quantities

The basketball starts from rest (\(v_i = 0\)) and reaches a speed of \(v_f = 6.6\) m/s at the bottom. We need to determine the height \(h\) of the hill for part (a).
02

Apply Conservation of Energy (Basketball)

Since we ignore friction, mechanical energy is conserved. The basketball's initial potential energy \(mgh\) is converted to kinetic energy \((1/2)mv^2\) at the bottom. Using the formula:\[ mgh = \frac{1}{2} mv_f^2\]Cancel the mass (\(m\)), and solve for \(h\):\[ gh = \frac{1}{2} v_f^2 \]Rearrange to find \(h\):\[ h = \frac{v_f^2}{2g} \]
03

Calculate the Height of the Hill

Substitute \(v_f = 6.6\) m/s and \(g = 9.8 \text{ m/s}^2\):\[ h = \frac{(6.6)^2}{2 \times 9.8} = \frac{43.56}{19.6} \approx 2.22 \text{ meters} \]So, the height of the hill is approximately \(2.22\) meters.
04

Determine the Translational Speed of the Frozen Juice Can

For the can of frozen juice, use the same principle of energy conservation. The height \(h\) is the same. Use the initial potential energy and set it equal to the final kinetic energy:\[ mgh = \frac{1}{2} mv_{f, \text{can}}^2 \]Similar to the basketball, mass \(m\) cancels out, yielding:\[ gh = \frac{1}{2} v_{f, \text{can}}^2\]
05

Solve for the Translational Speed of the Frozen Juice Can

Solve for \(v_{f, \text{can}}\):\[ v_{f, \text{can}} = \sqrt{2gh}\]We already found \(h = 2.22\) m, so substitute values with \(g = 9.8 \text{ m/s}^2\):\[ v_{f, \text{can}} = \sqrt{2 \times 9.8 \times 2.22} = \sqrt{43.56} \approx 6.6 \text{ m/s} \]The frozen juice can have the same translation speed of \(6.6\) m/s at the bottom as the basketball.

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Key Concepts

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

Mechanical Energy
Mechanical energy is the sum of both kinetic energy and potential energy in a system. In this specific hill scenario, the mechanical energy of objects like the basketball or the can of juice revolves around how their energy transforms as they roll down the hill.
Since friction is ignored in this exercise, the mechanical energy becomes conserved. That means:
  • The total energy remains the same at the start and at the bottom.
  • Potential energy starts high at the top and converts into kinetic energy at the bottom.
In simple terms, as an object rolls down the hill, all of its stored energy (potential energy) becomes motion energy (kinetic energy). Therefore, the sum of kinetic and potential energy stays constant for both the basketball and the can of juice. This principle of conservation of mechanical energy is key in solving such physics problems.
Kinetic Energy
Kinetic energy is the energy of motion. Any time an object is moving, it has kinetic energy. The faster it moves, the more kinetic energy it possesses. This is mathematically expressed by the formula:\[KE = \frac{1}{2}mv^2\]
where:
  • \(m\) is the mass of the object
  • \(v\) is its velocity.
The key to understanding this part of the exercise is realizing that as the basketball or the juice can rolls down the hill, potential energy is converted into kinetic energy. At the bottom, all the energy is in the form of kinetic energy. This situation shows how the velocity of an object connects directly to its energy. Hence, once you calculate potential energy at the top, determining the kinetic energy and thus the speed of the object at the bottom becomes straightforward.
Potential Energy
Potential energy can be thought of as stored energy that has the potential to do work. When an object is elevated at a height, like at the top of the hill in our problem, it has gravitational potential energy. The formula for this energy is:\[PE = mgh\]
where:
  • \(m\) is the mass of the object,
  • \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) on Earth),
  • \(h\) is the height above the ground.
This form of energy is all about position. The higher the object is lifted, the more potential energy it holds. For the hill example, the potential energy at the top is completely converted to kinetic energy at the bottom. By using the critical idea that potential energy translates to kinetic energy without loss, we can determine details like speed at the base or the estimation of the hill's height, as shown in the solved steps.

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Most popular questions from this chapter

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{s},\) and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s} .\) Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{m} .\) The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

A helicopter has two blades (see Figure 8.11 ); each blade has a mass of \(240 \mathrm{kg}\) and can be approximated as a thin rod of length \(6.7 \mathrm{m} .\) The blades are rotating at an angular speed of \(44 \mathrm{rad} / \mathrm{s}\). (a) What is the total moment of inertia of the two blades about the axis of rotation? (b) Determine the rotational kinetic energy of the spinning blades.

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+\) \(M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

The wheel of a car has a radius of \(0.350 \mathrm{m}\). The engine of the car applies a torque of \(295 \mathrm{N} \cdot \mathrm{m}\) to this wheel, which does not slip against the road surface. since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

The drawing shows an outstretched arm \((0.61 \mathrm{m}\) in length) that is parallel to the floor. The arm is pulling downward against the ring attached to the pulley system, in order to hold the \(98-N\) weight stationary. To pull the arm downward, the latissimus dorsi muscle applies the force \(\overrightarrow{\mathbf{M}}\) in the drawing, at a point that is \(0.069 \mathrm{m}\) from the shoulder joint and oriented at an angle of \(29^{\circ} .\) The arm has a weight of \(47 \mathrm{N}\) and a center of gravity (cg) that is located 0.28 \(\mathrm{m}\) from the shoulder joint. Find the magnitude of \(\overrightarrow{\mathbf{M}}\)

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