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Chapter 6: Problem 50

For sport, a \(12 \mathrm{~kg}\) armadillo runs onto a large pond of level, frictionless ice with an initial velocity of \(5.0 \mathrm{~m} / \mathrm{s}\) along the positive direction of an \(x\) axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of \(17 \mathrm{~N}\) in the positive direction of the \(y\) axis. In unit-vector notation, what are the animal's (a) velocity and (b) position vector when it has slid for \(3.0 \mathrm{~s}\) ?

Short Answer

Expert verified
Velocity vector: \( 5.0 \hat{i} + 4.26 \hat{j} \text{ m/s} \)Position vector: \( 15.0 \hat{i} + 6.39 \hat{j} \text{ m} \)

Step by step solution

01

Identify Given Information

First, note the given data for the problem: - Mass of the armadillo: 12 kg - Initial velocity: 5.0 m/s along the x-axis - Force by wind: 17 N along the y-axis - Time duration: 3.0 s
02

Calculate acceleration due to wind

Using Newton's second law, calculate the acceleration caused by the wind:a = \( \frac{F}{m} \) where F = 17 N and m = 12 kg.a = \( \frac{17 \text{ N}}{12 \text{ kg}} \)a = 1.42 m/s虏 along the y-axis.
03

Determine Velocity in the x-direction

As there is no force acting in the x-direction, the velocity remains constant due to the frictionless ice:\( v_{x} = 5.0 \text{ m/s} \)
04

Determine Velocity in the y-direction

Use the kinematic equation to find the final velocity in the y-direction:v = u + at where u = 0 m/s, a = 1.42 m/s虏, and t = 3.0 s.\( v_{y} = 0 + (1.42 \text{ m/s}^2 \cdot 3.0 \text{ s}) \)\( v_{y} = 4.26 \text{ m/s} \)
05

Write the Velocity Vector in Unit Notation

Now, combine the x and y components to write the velocity vector:\( \vec{v} = 5.0 \hat{i} + 4.26 \hat{j} \text{ m/s} \)
06

Determine Position in x-direction

Use the kinematic equation to find the position in the x-direction:s = ut + \frac{1}{2}at^2 where u = 5.0 m/s, a = 0, and t = 3.0 s.\( x = 5.0 \text{ m/s} \cdot 3.0 \text{ s} \)\( x = 15.0 \text{ m} \)
07

Determine Position in y-direction

Use the kinematic equation to find the position in the y-direction:s = ut + \frac{1}{2}at^2 where u = 0 m/s, a = 1.42 m/s虏, and t = 3.0 s.\( y = 0 + \frac{1}{2} \cdot 1.42 \text{ m/s}^2 \cdot (3.0 \text{ s})^2 \)\( y = 6.39 \text{ m} \)
08

Write the Position Vector in Unit Notation

Now, combine the x and y components to write the position vector:\( \vec{r} = 15.0 \hat{i} + 6.39 \hat{j} \text{ m} \)

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Key Concepts

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

kinematics equations
Where:
  • \(u\) is the initial velocity
  • \(v\) is the final velocity
  • \(a\) is the acceleration
  • \(t\) is the time taken
  • \(s\) is the displacement
In the problem, we determined the armadillo's velocity and position using these equations. For example, to find the y-component of velocity after 3 seconds, we substituted the values into the equation \(v = 0 + (1.42 \, \text{m/s}^2 \cdot 3.0 \, \text{s}) \), resulting in a final velocity of 4.26 m/s.
acceleration
Using Newton's second law, we calculate acceleration as follows:
\( a = \frac{F}{m} = \frac{17 \, \text{N}}{12 \, \text{kg}} = 1.42 \, \text{m/s}^2 \).

This acceleration causes a change in the armadillo's velocity in the y-direction over time. Since there's no force in the x-direction, the velocity in that direction remains constant.
unit vector notation
Combine the components to form a vector.
  • The velocity vector is \( \vec{v} = 5.0 \hat{i} + 4.26 \hat{j} \, \text{m/s}\)
  • The position vector is \( \vec{r} = 15.0 \hat{i} + 6.39 \hat{j} \, \text{m}\)
Unit vector notation helps in precisely describing multidimensional motion in physics problems. It provides both magnitude and direction clearly, making it easier to analyze and understand different components of motion.
frictionless motion
This results in the armadillo's x-velocity remaining constant. The only force acting on the armadillo is the wind force in the y-direction, causing an acceleration and changing the y-velocity.

This simplifies the analysis as we don't have to account for energy lost to friction. It helps us focus on the core concepts of forces and motion, making calculations straightforward and helping to deepen understanding of the underlying principles.

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Most popular questions from this chapter

Luggage is transported from one location to another in an airport by a conveyor belt. At a certain location, the belt moves down an incline that makes an angle of \(2.5^{\circ}\) with the horizontal. Assume that with such a slight angle there is no slipping of the luggage. Determine the magnitude and direction of the frictional force by the belt on a box weighing \(69 \mathrm{~N}\) when the box is on the inclined portion of the belt for the following situations: (a) The belt is stationary. (b) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is constant. (c) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is increasing at a rate of \(0.20 \mathrm{~m} / \mathrm{s}^{2}\). (d) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is decreasing at a rate of \(0.20 \mathrm{~m} / \mathrm{s}^{2}\). (e) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is increasing at a rate of \(0.57 \mathrm{~m} / \mathrm{s}^{2}\).

A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. (a) What is the magnitude of the net force acting on the rider? (b) What is the magnitude of the force on the rider from the motorcycle?

During an Olympic bobsled run, the Jamaican team makes a turn of radius \(7.6 \mathrm{~m}\) at a speed of \(96.6 \mathrm{~km} / \mathrm{h}\). What is their acceleration in \(g\) -units? \((1 g\) -unit \(=\) \(\left.9.8 \mathrm{~m} / \mathrm{s}^{2} .\right)\)

calculate the ratio of the drag force on a passenger jet flying with a speed of \(1000 \mathrm{~km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a prop-driven transport flying at half the speed and half the altitude of the jet. At \(10 \mathrm{~km}\) the density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\), and at \(5.0 \mathrm{~km}\) it is \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\). Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient \(C\).

A worker pushes horizontally on a \(35 \mathrm{~kg}\) crate with a force of magnitude \(110 \mathrm{~N}\). The coefficient of static friction between the crate and the floor is \(0.37\). (a) What is the frictional force on the crate from the floor? (b) What is the maximum magnitude \(f_{\max }^{\text {stat }}\) of the static frictional force under the circumstances? (c) Does the crate move? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker's \(110 \mathrm{~N}\) push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?
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