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Chapter 8: Problem 38

A block of ice with a mass of 2.50 \(\mathrm{kg}\) is moving on a fric- tionless, 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 s: (a) a force of 5.00 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

Understand the initial conditions

The block of ice has an initial velocity \(v_0 = 8.00\, \mathrm{m/s}\) to the right, and a mass \(m = 2.50\, \mathrm{kg}\). The initial velocity is positive, as it moves to the right.
02

Calculate acceleration for part (a)

For the force of 5.00 \(\mathrm{N}\) directed to the right, use Newton's second law, \( F = ma \). Thus, \( a = \frac{F}{m} = \frac{5.00\, \mathrm{N}}{2.50\, \mathrm{kg}} = 2.00\, \mathrm{m/s^2} \).
03

Determine the final velocity for part (a)

Using the equation for velocity \( v = v_0 + at \), substitute \( v_0 = 8.00\, \mathrm{m/s}, a = 2.00\, \mathrm{m/s^2}, \) and \( t = 5.00\, \mathrm{s} \): \( v = 8.00 + (2.00)(5.00) = 18.00\, \mathrm{m/s} \). The final velocity after the force is applied is 18.00 \(\mathrm{m/s}\) to the right.
04

Calculate acceleration for part (b)

For the force of 7.00 \(\mathrm{N}\) directed to the left, use Newton's second law reverse direction, \( a = \frac{-7.00\, \mathrm{N}}{2.50\, \mathrm{kg}} = -2.80\, \mathrm{m/s^2} \).
05

Determine the final velocity for part (b)

Using the equation for velocity \( v = v_0 + at \), substitute \( v_0 = 8.00\, \mathrm{m/s}, a = -2.80\, \mathrm{m/s^2}, \) and \( t = 5.00\, \mathrm{s} \): \( v = 8.00 + (-2.80)(5.00) = -6.00\, \mathrm{m/s} \). 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 the branch of physics that deals with the motion of objects without considering the causes of this motion. When you're studying kinematics, you're looking at how objects move鈥攕omething you could describe with graphs, equations, or words. A common approach is to determine velocity, acceleration, and displacement over a given period.

In the context of our exercise, kinematics helps describe how the ice block's velocity changes over time. The initial velocity, time of travel, and any forces acting upon it directly influence how these kinematic variables evolve. With forces acting, the block either speeds up or slows down depending upon the direction of force relative to its motion.

By knowing the initial conditions (initial velocity, direction, and forces), you can use kinematic equations to predict what happens to the object later in its journey.
Force and Motion
Force and motion are key concepts that explain why objects move and how they change their motion. According to Newton's Second Law, an object will only accelerate if there is a net force acting upon it. This law is often summed up as \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration.
  • In the exercise, forces are applied to the ice block to alter its velocity.
  • A 5 N force to the right results in acceleration in the same direction, while a 7 N force to the left changes the direction of acceleration.
Understanding how forces impact an object's motion can help predict outcomes like the ice block speeding up, slowing down, or changing direction. The resulting acceleration from these forces is crucial for determining changes in velocity.
Velocity Calculation
Velocity is more than just speed; it's speed with direction. Calculating velocity involves knowing the initial velocity, the time over which the force acts, and the acceleration resulting from that force.

To arrive at the final velocity, we use the formula \( v = v_0 + at \), where \( v_0 \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time period the force is applied.
  • For part (a), the initial velocity is 8.00 m/s, the acceleration is 2.00 m/s虏, and the force is applied for 5.00 seconds. This gives a final velocity of 18.00 m/s to the right.
  • However, for part (b), with a force to the left, the acceleration becomes -2.80 m/s虏, resulting in a final velocity of -6.00 m/s, indicating movement to the left.
So, by plugging these values into the velocity formula, you can determine exactly how the velocity changes in different scenarios.

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Most popular questions from this chapter

\(\bullet\) Detecting planets around other stars. Roughly 500 planets have so far been detected beyond our solar system. This is accomplished by looking for the effect the planet has on the star. The star is not truly stationary; instead, it and its planets orbit around the center of mass of the system. Astronomers can measure this wobble in the position of a star.(a) For a star with the mass and size of our sun and having a planet with five times the mass of Jupiter, where would the center of mass of this system be located, relative to the center of the star, if the distance from the star to the planet was the same as the distance from Jupiter to our sun? (Consult Appendix E.) (b) If the planet had earth's mass, where would the center of mass of the system be located if the planet was just as far from the star as the earth is from the sun? (c) In view of your results in parts (a) and (b), why is it much easier to detect stars having large planets rather than small ones?

\(\bullet\) On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a frozen lake. With your finger, you flick the penny toward the nickel with a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) . The coins collide elastically; calculate both their final velocities (speed and direction).

Forensic science. Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure 8.49.) The meas ured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet into a 2.50 kg block hanging by a thin wire 75.0 \(\mathrm{cm}\) long. causing the block to swing upward to a maximum angle of \(34.7^{\circ}\) from the vertical. What was the original speed of this bullet?

Recoil speed of the earth. In principle, any time someone jumps up, the earth moves in the opposite direction. To see why we are unaware of this motion, calculate the recoil speed of the earth when a 75 kg person jumps upward at a speed of 2.0 \(\mathrm{m} / \mathrm{s} .\) Consult Appendix E as needed.

\(\cdot\) 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 neglected. (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?
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