/*! 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 53 A cement block accidentally fall... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 53

A cement block accidentally falls from rest from the ledge of a 53.0 -m-high building. When the block is \(14.0 \mathrm{~m}\) above the ground, a man, \(2.00 \mathrm{~m}\) tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

Short Answer

Expert verified
The man has approximately 1.691 seconds to move out of the way.

Step by step solution

01

Determine the Initial Conditions

The block falls from a height of 53.0 m. Its initial velocity is 0 m/s since it was at rest. We are to find the time it takes to fall from a height of 14.0 m above the ground to the ground level where the man is standing.
02

Apply the Kinematic Equation

Use the kinematic equation for one-dimensional motion under constant acceleration to determine the time \( t \) it takes for the block to fall the last 14.0 m. The equation is: \[ h = v_0 t + \frac{1}{2} a t^2 \]where:- \( h = 14.0 \) m (distance fallen);- \( v_0 = 0 \) m/s (initial velocity);- \( a = 9.81 \) m/s虏 (acceleration due to gravity).Substituting the values, we get:\[ 14.0 = 0 \cdot t + \frac{1}{2} \cdot 9.81 \cdot t^2 \]
03

Solve for Time

Rearrange and solve the equation from Step 2:\[ 14.0 = \frac{1}{2} \cdot 9.81 \cdot t^2 \]\[ 28.0 = 9.81 \cdot t^2 \]\[ t^2 = \frac{28.0}{9.81} \]\[ t^2 = 2.854 \]\[ t = \sqrt{2.854} \]\[ t \approx 1.691 \text{ seconds} \]
04

Conclusion

The block takes approximately 1.691 seconds to fall from 14.0 m to the ground. Therefore, the man has roughly 1.691 seconds to move out of the way to avoid the falling block.

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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 vital in solving problems related to motion, particularly when dealing with objects under acceleration. They relate an object's initial velocity, final velocity, acceleration, time, and displacement. For objects that start from rest, like the cement block in this exercise, the initial velocity (\(v_0\)) is zero.
Understanding kinematic equations can help you calculate not just how far an object will go, but also how long it will take to get there, and at what speed it will travel.
A few common kinematic equations include:
  • \(v = v_0 + at\) - Final velocity after time \(t\)
  • \(s = v_0t + \frac{1}{2}at^2\) - Displacement after time \(t\)
  • \(v^2 = v_0^2 + 2as\) - Relates velocity and displacement
In our exercise, we used the second kinematic equation to find time because we knew the displacement and acceleration.
Constant Acceleration
Constant acceleration refers to the situation where an object's velocity changes at a consistent rate over time. This means the object's acceleration remains the same both in magnitude and direction. In the given exercise, the cement block experiences constant acceleration as it falls due to gravity.
For constant acceleration, you can expect the equations of motion to simplify since acceleration (\(a\)) doesn't change. This simplifies calculations significantly.
In everyday physics problems, gravity often represents constant acceleration because its value doesn't vary considerably over short distances. Therefore, using the value 9.81 m/s虏 for gravitational acceleration in calculations is common practice.
Acceleration Due to Gravity
Acceleration due to gravity is the rate at which an object accelerates when falling freely under the influence of gravitational force alone.
This acceleration is usually denoted by \(g\) and its standard value is approximately 9.81 m/s虏 on Earth's surface, which remains relatively constant.
Because the force of gravity acts uniformly on all objects regardless of their mass, any object under free fall in a vacuum near Earth's surface will have the same acceleration due to gravity.
In the exercise, this constant acceleration value allows us to calculate how long it takes for the cement block to travel the final 14 meters to the ground.
One-Dimensional Motion
One-dimensional motion refers to movement along a straight line, where an object can only move forward or backward. In such scenarios, we only need one coordinate, usually denoted by \(x\) or \(s\), to describe the object's motion.
When analyzing such motion, factors like velocity, acceleration, and displacement are calculated within this linear path. The motion of the cement block dropping straight down aligns with this concept.
One-dimensional problems are simpler to solve because they involve basic arithmetic. They focus on changes in position along a single axis, making them a great starting point to learn about more complex multidimensional motion problems.

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Most popular questions from this chapter

You are on a train that is traveling at \(3.0 \mathrm{~m} / \mathrm{s}\) along a level straight track. Very near and parallel to the track is a wall that slopes upward at a \(12^{\circ}\) angle with the horizontal. As you face the window \((0.90 \mathrm{~m}\) high, \(2.0 \mathrm{~m}\) wide ) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner \(\mathrm{A}\) and eventually disappears at window corner \(\mathrm{B}\). How much time passes between appearance and disappearance of the upper edge of the wall?

While standing on a bridge \(15.0 \mathrm{~m}\) above the ground, you drop a stone from rest. When the stone has fallen \(3.20 \mathrm{~m}\), you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.

Review Interactive Solution 2.49 at before beginning this problem. A woman on a bridge \(75.0 \mathrm{~m}\) high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 7.00 \(\mathrm{m}\) more to travel before passing under the bridge. The stone hits the water \(4.00 \mathrm{~m}\) in front of the raft. Find the speed of the raft.

Two cars cover the same distance in a straight line. Car \(\mathrm{A}\) covers the distance at a constant velocity. Car B starts from rest, maintains a constant acceleration, and covers the distance in the same time as car A. (a) Is car B's final velocity smaller than, equal to, or greater than car A's constant velocity? Explain. (b) Is car B's average velocity smaller than, equal to, or greater than car A's constant velocity? Why? (c) Given a value for the time to cover the distance and the fact that car B starts from rest, what other variable must be known before car B's acceleration can be determined? Provide a reason for your answer. Problem Both cars cover a distance of \(460 \mathrm{~m}\) in \(210 \mathrm{~s}\). Assume that they are moving in the \(+x\) direction. Determine (a) the constant velocity of car \(\mathrm{A},\) (b) the final velocity of car \(\mathrm{B},\) and \((\mathrm{c})\) the acceleration of car \(\mathrm{B}\). Verify that your answers are consistent with your answers to the Concept Questions.

In preparation for this problem, review Conceptual Example 7 . From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The initial speed of the pebble is \(9.0 \mathrm{~m} / \mathrm{s}\). (a) What is the acceleration (magnitude and direction) of the pebble during the downward motion? Is the pebble decelerating? Explain. (b) After \(0.50 \mathrm{~s}\), how far beneath the cliff top is the pebble?
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