/*! 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 105 A 2.0-kg object has a velocity o... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 105

A 2.0-kg object has a velocity of 4.0 铆 m/s at \(t=0 . \quad\) A constant resultant force of \((2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) \mathrm{N}\) then acts on the object for 3.0 s. What is the magnitude of the object's velocity at the end of the 3.0 -s interval?

Short Answer

Expert verified
The magnitude of the object's velocity at the end of the 3.0-s interval is approximately \(9.22 m/s\).

Step by step solution

01

Determine the initial velocity components on x and y axes

The initial velocity of the object is given as \(4.0 \hat{\mathbf{i}} m/s\). This means that the object has an initial horizontal velocity, and its x-component (Vx) is 4.0 m/s. The initial y-component (Vy) can be considered to be 0 m/s since there's no vertical motion mentioned. Initial velocity: Vx = 4.0 m/s (in the x-axis) Vy = 0 m/s (in the y-axis)
02

Calculate the acceleration components on x and y axes

The force acting on the object is given as \((2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) N\). The mass of the object is 2.0 kg. We can find the acceleration of the object using Newton's second law (F = ma). X-axis acceleration (ax): \(a_x = \frac{F_x}{m} = \frac{2.0 N}{2.0 kg} = 1.0 m/s^2\) Y-axis acceleration (ay): \(a_y = \frac{F_y}{m} = \frac{4.0 N}{2.0 kg} = 2.0 m/s^2\)
03

Calculate the final velocity components on x and y axes after 3.0s

The constant force acts on the object for a time of 3.0s. We can find the final velocity of the object using the acceleration components and the initial velocities: Final x-component velocity (Vfx): \(V_{fx} = V_{x(i)} + a_x \cdot t = 4.0 m/s + (1.0 m/s^2)(3.0 s) = 7.0 m/s\) Final y-component velocity (Vfy): \(V_{fy} = V_{y(i)} + a_y \cdot t = 0 m/s + (2.0 m/s^2)(3.0 s) = 6.0 m/s\)
04

Calculate the magnitude of the object's velocity at the end

We have the final velocity components on x and y axes, and we can now calculate the magnitude of the final velocity using the Pythagorean theorem: \(V_f = \sqrt{V_{fx}^2 + V_{fy}^2} = \sqrt{(7.0 m/s)^2 + (6.0 m/s)^2} = \sqrt{49 + 36 }\) \(V_f = \sqrt{85} \approx 9.22 m/s\) The magnitude of the object's velocity at the end of the 3.0-s interval is approximately 9.22 m/s.

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Key Concepts

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

Understanding Vector Components
Vectors are essential when dealing with forces and motion in physics. In this exercise, we look at the components of velocity as well as force, which are both vector quantities. To simplify calculations, vectors are often broken down into their components along the x and y axes. Understanding these components allows us to apply Newton's Second Law effectively. For an object initially moving with a velocity \(4.0 \hat{i} \text{ m/s}\), we see that:
  • The horizontal component of velocity (x-axis) is 4.0 m/s.
  • The initial vertical component of velocity (y-axis) is 0 m/s, as there's no vertical movement provided initially.
Breaking vectors into components on each axis helps manage and calculate their effect on an object. This approach makes predicting the object's movement much easier.
Acceleration Calculation and Newton's Second Law
Newton's Second Law of Motion tells us that force equals mass times acceleration (\(F = ma \)). In this exercise, it helps us find how an object's velocity changes over time when an external force acts on it.We use the force vector \((2.0 \hat{i} + 4.0 \hat{j}) \text{ N}\) to find acceleration:
  • X-axis component of acceleration: \(a_x = \frac{2.0 \text{ N}}{2.0 \text{ kg}} = 1.0 \text{ m/s}^2\)
  • Y-axis component of acceleration: \(a_y = \frac{4.0 \text{ N}}{2.0 \text{ kg}} = 2.0 \text{ m/s}^2\)
The calculation of these components is straightforward. It involves dividing the respective force component by the object's mass. With these values, we understand how the object accelerates in each direction.
Final Velocity Determination
Once we calculate the acceleration, we can determine how the velocity of the object changes over a given period, in this case, 3.0 seconds. We apply the formula for velocity in each dimension using the initial velocity and the calculated acceleration:Final velocities:
  • X-component: \(V_{fx} = V_{x(i)} + a_x \cdot t = 4.0 \text{ m/s} + (1.0 \text{ m/s}^2)(3.0 \text{ s}) = 7.0 \text{ m/s} \)
  • Y-component: \(V_{fy} = V_{y(i)} + a_y \cdot t = 0 \text{ m/s} + (2.0 \text{ m/s}^2)(3.0 \text{ s}) = 6.0 \text{ m/s}\)
To find the magnitude of the final velocity vector, use the Pythagorean theorem: \[ V_f = \sqrt{V_{fx}^2 + V_{fy}^2} = \sqrt{(7.0 \text{ m/s})^2 + (6.0 \text{ m/s})^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.22 \text{ m/s} \]Thus, the object鈥檚 velocity magnitude at the end of the interval is approximately 9.22 m/s, giving a full picture of how it moves after 3 seconds.

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Most popular questions from this chapter

A basketball player jumps straight up for a ball. To do this, he lowers his body \(0.300 \mathrm{m}\) and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him \(0.900 \mathrm{m}\) above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of \(0.300 \mathrm{m}\). (c) Calculate the force he exerts on the floor to do this, given that his mass is \(110.0 \mathrm{kg}\).

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