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Chapter 7: Problem 45

A particle of mass \(2 \mathrm{~kg}\) moves with an initial velocity of \(v=(4 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}\). A constant force of \(F=-20 \hat{j} \mathrm{~N}\) is applied on the particle. Initially, the particle was at (0, 0). The \(x\) -coordinate of the particle when its \(y\) -coordinate again becomes zero is given by a. \(1.2 \mathrm{~m}\) b. \(4.8 \mathrm{~m}\) c. \(6.0 \mathrm{~m}\) d. \(3.2 \mathrm{~m}\)

Short Answer

Expert verified
The x-coordinate is 3.2 m, corresponding to option (d).

Step by step solution

01

Analyze the Given Information

We have a particle with mass \(m = 2 \text{ kg}\), and initial velocity \(v = \langle 4 \hat{i}, 4 \hat{j} \rangle \text{ m/s}\). A constant force \(F = \langle 0, -20 \hat{j} \rangle \text{ N}\) is applied to the particle. Initially, the particle is at position \((0, 0)\). The problem asks for the \(x\)-coordinate when the \(y\)-coordinate becomes zero again.
02

Determine the Acceleration

Use Newton's second law to find the acceleration. The mass of the particle is \(2 \text{ kg}\), and the force is \(\langle 0, -20 \rangle \text{ N}\). Thus, acceleration \(a\) is given by: \[a = \frac{F}{m} = \frac{\langle 0, -20 \hat{j} \rangle}{2} = \langle 0, -10 \hat{j} \rangle \text{ m/s}^2.\]
03

Set Up Equations of Motion for the y-Direction

Use the kinematic equation for the \(y\)-direction: \[y = v_{0y} t + \frac{1}{2} a_y t^2,\] where \(v_{0y} = 4 \text{ m/s}\) and \(a_y = -10 \text{ m/s}^2\). We need \(y = 0\) again, when it was already 0 initially, so:\[0 = 4t - 5t^2.\]
04

Solve for Time When y = 0

The equation from Step 3 is: \[0 = 4t - 5t^2.\]Factor out \(t\):\[t(4 - 5t) = 0.\]This gives solutions \(t = 0\) (initial position) and \(t = \frac{4}{5} \text{ s}\).We need \(t = \frac{4}{5} \text{ s}\) for when \(y = 0\) again.
05

Use Equations of Motion for the x-Direction

Since there is no force in the x-direction, the velocity \(v_{0x} = 4 \text{ m/s}\) remains constant. The position of the particle in the \(x\)-direction is given by:\[x = v_{0x} t = 4 \times \frac{4}{5} = \frac{16}{5} = 3.2 \text{ m}.\]
06

Identify the Correct Option

Compare the calculated \(x\)-coordinate to the given options. The \(x\)-coordinate where the \(y\)-coordinate becomes zero again is \(3.2\text{ m}\), which corresponds to option (d).

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools in physics that help describe the motion of objects. These equations relate the five key variables of motion: displacement, initial velocity, final velocity, acceleration, and time. Why are they important? Because they allow us to analyze different motion scenarios without direct experimentation. In the case of our problem, we used the kinematic equation to solve for the displacement in the y-direction:
  • Initial velocity (20y): 4 m/s
  • Acceleration (ay): -10 m/s虏, due to the constant force acting opposite to the motion
  • Our task was to find when the position in the y-direction becomes zero again.
By simplifying the equation, we determine two time values. The initial moment at t = 0 and later when the y-coordinate again becomes zero. This helped pinpoint the duration of motion needed to calculate displacement in the x-direction.
Newton's Second Law
Newton's Second Law is crucial for understanding how forces affect motion. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass, or simply put, \(F = ma\). In our example, this law helped determine the acceleration:
  • Force applied (F): -20 N
  • Mass (m): 2 kg
By applying Newton's Second Law, we calculated the acceleration of the particle in the y-direction. Since the force is constant and acts in the negative y-direction, the acceleration comes out to be -10 m/s虏. Knowing this value is essential, as it directly influences how the particle's velocity changes over time.
Constant Force
Constant force plays a significant role in predicting motion behavior. When a force of constant magnitude and direction acts on an object, it results in unchanging acceleration. This concept allows for simpler calculations using kinematic and dynamic equations. In our scenario, the constant force of -20 N, acting solely in the y-direction, shows its effect by:
  • Providing a steady acceleration, which changes only the velocity in the y-direction.
  • Ensuring no change in the x-direction, as no force is applied there.
This consistency allows us to easily determine the particle's position over a set time period, using simple linear projection.
Two-Dimensional Motion
The analysis of two-dimensional motion involves understanding movement along both the x and y axes. Objects traveling in two dimensions exhibit unique behaviors characteristic of projectile motion. In this current example, the particle starts with an initial velocity vector (4 i + 4 j) m/s, which implies:
  • Moving diagonally at first, with constant speed in the x-direction since no force acts there
  • Changing velocity in the y-direction due to the constant force applied
By solving for when the y-component of position again equals zero, we reunited both dimensions under a calculated time of 4/5 s. This allowed accurate tracking of the x-position, remaining unaffected by forces but consistently reliant on time to replicate projectile trajectory.

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Most popular questions from this chapter

The masses of the blocks \(A\) and \(B\) are \(m\) and \(M .\) Between \(A\) and \(B\) there is a constant frictional force \(F\), and \(B\) can slide frictionlessly on horizontal surface (see Fig. \(7.378\) ). \(A\) is set in motion with velocity while \(B\) is at rest. What is the distance moved by \(A\) relative to \(B\) before they move with the same velocity? a. \(\frac{m M v_{0}^{2}}{F(m-M)}\) b. \(\frac{m M v_{0}^{2}}{2 F(m-M)}\) c. \(\frac{m M v_{0}^{2}}{F(m+M)}\) d. \(\frac{m M v_{0}^{2}}{2 F(m+M)}\)

Two skaters weighing in the ratio \(4: 5\) and \(9 \mathrm{~m}\) apart are skating on a smooth frictionless surface. They pull on a rope stretched between them. The ratio of the distance covered by them when they meet each other will be a. \(5: 4\) b. \(4: 5\) c. \(25: 16\) d. \(16: 25\)

A block of mass \(1 \mathrm{~kg}\) is at rest on a horizontal table. The coefficient of static friction between the block and the table is \(0.50\). If \(g=10 \mathrm{~ms}^{-2}\), then the magnitude of a force acting upwards at an angle of \(60^{\circ}\) from the horizontal that will just start the block moving is: a. \(5 \mathrm{~N}\) b. \(5.36 \mathrm{~N}\) c. \(74.6 \mathrm{~N}\) d. \(10 \mathrm{~N}\)

A horizontal force of \(10 \mathrm{~N}\) is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is \(0.2\) (Fig. 7.349). The weight of the block is a. \(2 \mathrm{~N}\) b. \(20 \mathrm{~N}\) c. \(50 \mathrm{~N}\) d. \(100 \mathrm{~N}\)

The upper half of an inclined plane with inclination \(\phi\) is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by a. \(2 \tan \phi\) b. \(\tan \phi\) c. \(2 \sin \phi\) d. \(2 \cos \phi\)
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