/*! 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 35 A block of ice with a mass of \(... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 35

A block of ice with a mass of \(2.50 \mathrm{~kg}\) is moving on a frictionless, horizontal surface. At \(t=0,\) the block is moving to the right with a velocity of magnitude \(8.00 \mathrm{~m} / \mathrm{s}\). Calculate the magnitude and direction of the velocity of the block after each of the following forces has been applied for \(5.00 \mathrm{~s}\) : (a) a force of \(5.00 \mathrm{~N}\) directed to the right; (b) a force of \(7.00 \mathrm{~N}\) directed to the left.

Short Answer

Expert verified
(a) 18.00 m/s to the right; (b) 6.00 m/s to the left.

Step by step solution

01

Calculate Acceleration for Force (a)

First, find the acceleration imparted by the force of 5.00 N to the right. Use Newton's second law, given by the formula: \[ F = ma \]Solving for acceleration \( a \), we have:\[ a = \frac{F}{m} = \frac{5.00 \text{ N}}{2.50 \text{ kg}} = 2.00 \text{ m/s}^2 \].
02

Calculate Final Velocity for Force (a)

Next, calculate the final velocity of the block after 5.00 s, using the equation for uniformly accelerated motion:\[ v = u + at \]Where:- \( v \) is the final velocity- \( u = 8.00 \text{ m/s} \) is the initial velocity- \( a = 2.00 \text{ m/s}^2 \) is the acceleration- \( t = 5.00 \text{ s} \) is the timePlugging in the values:\[ v = 8.00 \text{ m/s} + (2.00 \text{ m/s}^2 \times 5.00 \text{ s}) = 18.00 \text{ m/s} \].The velocity is directed to the right.
03

Calculate Acceleration for Force (b)

For the force of 7.00 N directed to the left, first find the acceleration:\[ a = \frac{F}{m} = \frac{7.00 \text{ N}}{2.50 \text{ kg}} = 2.80 \text{ m/s}^2 \].Note that the acceleration is to the left because the force is to the left.
04

Calculate Final Velocity for Force (b)

Find the final velocity after applying the leftward force for 5.00 s:\[ v = u + at \]Where the values are:- \( u = 8.00 \text{ m/s} \)- \( a = -2.80 \text{ m/s}^2 \) (negative because the acceleration is in the opposite direction)- \( t = 5.00 \text{ s} \)Enter the known values:\[ v = 8.00 \text{ m/s} + (-2.80 \text{ m/s}^2 \times 5.00 \text{ s}) = -6.00 \text{ m/s} \]. The negative sign indicates the direction is to the left.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that studies motion without considering the forces that cause this motion. It focuses on describing motion in terms of objects moving from one point to another. The key elements of kinematics include displacement, velocity, and acceleration, providing crucial insights into how objects move.
In our exercise, the block of ice on a frictionless horizontal surface is a perfect illustration of a kinematics scenario. The initial conditions include:
  • Initial velocity: the speed and direction of the block at the start, which is given as 8.00 m/s to the right.
  • Time duration: the period during which forces are applied, noted as 5 seconds.
Understanding kinematics helps us predict future motion based on these initial conditions.
Velocity Calculation
Velocity is a vector quantity, meaning it includes both magnitude and direction. In the context of our exercise, calculating the final velocity of the block after a force is applied involves understanding how speed and direction change over time.
The formula used is:
\( v = u + at \),where:
  • \( v \) is the final velocity that we are looking to calculate.
  • \( u \) is the initial velocity, which is set at 8.00 m/s.
  • \( a \) is the acceleration caused by the applied force.
  • \( t \) is the time during which the force is applied, noted as 5.00 seconds.
This equation allows us to incorporate both the change in speed and the direction the block is moving in over time.
Acceleration
Acceleration is the rate of change of velocity with respect to time. It indicates how quickly an object is speeding up or slowing down. According to Newton's Second Law, acceleration is caused by a net force acting upon an object, and is calculated using the formula:
\[ a = \frac{F}{m} \],where:
  • \( F \) is the force applied.
  • \( m \) is the mass of the object.
The sign of acceleration depends on the direction of the force. In our exercise:
  • For a 5.00 N force to the right: This results in a positive acceleration of 2.00 m/s虏.
  • For a 7.00 N force to the left: This results in a negative acceleration of 2.80 m/s虏, reflecting a change in direction.
These calculations demonstrate how different forces lead to varying accelerations, impacting the object's velocity.

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Most popular questions from this chapter

A 4.25 g bullet traveling horizontally with a velocity of magnitude \(375 \mathrm{~m} / \mathrm{s}\) is fired into a wooden block with mass \(1.12 \mathrm{~kg},\) initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to \(122 \mathrm{~m} / \mathrm{s} .\) How fast is the block moving just after the bullet emerges from it?

A \(15.0 \mathrm{~kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (See Figure \(8.40 .\) ) Suddenly it is struck by a \(3.00 \mathrm{~kg}\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts the collision and the behavior after the collision -and apply the appropriate conservation law to each part.)

Three identical boxcars are coupled together and are moving at a constant speed of \(20.0 \mathrm{~m} / \mathrm{s}\) on a level track. They collide with another identical boxcar that is initially at rest and couple to it, so that the four cars roll on as a unit. Friction is small enough to be ignored. (a) What is the speed of the four cars? (b) What percentage of the kinetic energy of the boxcars is dissipated in the collision? What happened to this energy?

Cart \(A\) has a mass of \(5 \mathrm{~kg}\) and is moving in the \(+x\) direction at \(2 \mathrm{~m} / \mathrm{s}\). Cart \(B\) has a mass of \(2 \mathrm{~kg}\) and is moving in the \(+y\) direction at \(5 \mathrm{~m} / \mathrm{s}\). (a) Do the two carts have the same momentum? Explain. (b) Is the magnitude of the momentum of each cart the same? Explain. (c) Is the kinetic energy of each cart the same? Explain.

Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the water's surface to knock them into the water, where the fish can eat them. \(\mathrm{A} 65 \mathrm{~g}\) fish at rest just on the water's surface can expel a \(0.30 \mathrm{~g}\) drop of water in a short burst of \(5.0 \mathrm{~ms}\). High-speed measurements show that the water has a speed of \(2.5 \mathrm{~m} / \mathrm{s}\) just after the archerfish expels it. What is the speed of the archerfish immediately after it expels the drop of water? A. \(0.0025 \mathrm{~m} / \mathrm{s}\) B. \(0.012 \mathrm{~m} / \mathrm{s}\) C. \(0.75 \mathrm{~m} / \mathrm{s}\) D. \(2.5 \mathrm{~m} / \mathrm{s}\)
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