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Chapter 2: Problem 52

A bolt is dropped from a bridge under construction, falling \(100 \mathrm{~m}\) to the valley below the bridge. (a) In how much time does it pass through the last \(20 \%\) of its fall? What is its speed (b) when it begins that last \(20 \%\) of its fall and (c) when it reaches the valley beneath the bridge?

Short Answer

Expert verified
(a) 2.02 s, (b) 39.6 m/s, (c) 44.3 m/s.

Step by step solution

01

- Determine the total distance and its 80%

The total distance fallen by the bolt is 100 m. Since we are interested in the last 20% of the fall, we first calculate 80% of the 100 m to find the point at which this last section begins.Calculate 80% of 100 m:\[ 0.8 \times 100 = 80 \text{ m} \]Thus, the last 20% of the fall starts at 80 m from the top.
02

- Calculate time for the last 20% of the fall

The bolt falls freely under gravity, starting from rest. We use the kinematic equation to determine the time it takes to fall the last 20 m (from 80 m to 100 m):\[ s = ut + \frac{1}{2}gt^2 \]For the last 20 m:\[ 20 = 0 + \frac{1}{2}(9.8)t^2 \]Solving for \(t\), we get:\[ 20 = 4.9t^2 \]\[ t^2 = \frac{20}{4.9} \]\[ t^2 \approx 4.08 \]\[ t \approx 2.02 \text{ seconds} \]
03

- Find its speed at the 80 m mark

We use the kinematic equation \( v^2 = u^2 + 2as \) to find the speed when the bolt has fallen 80 m:\[ v^2 = 0 + 2 \times 9.8 \times 80 \]\[ v^2 = 1568 \]\[ v = \sqrt{1568} \approx 39.6 \text{ m/s} \]Thus, the speed at 80 m is approximately 39.6 m/s.
04

- Calculate the speed at the bottom

Now, let's find the speed when the bolt reaches the bottom of the 100 m fall using the same equation:\[ v^2 = u^2 + 2as \]Since the total fall is 100 m, \( a = 9.8 \text{ m/s}^2 \), and starting from rest \( u = 0 \):\[ v^2 = 2 \times 9.8 \times 100 \]\[ v^2 = 1960 \]\[ v = \sqrt{1960} \approx 44.3 \text{ m/s} \]The speed as the bolt reaches the valley is approximately 44.3 m/s.

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Key Concepts

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

Free Fall
Free fall is a simple yet fascinating concept in physics involving objects moving under the sole influence of gravity. This means that when an object is in free fall, no other forces, such as air resistance, affect its motion. In our problem, the bolt is in free fall from a bridge. Because of gravity, it accelerates downwards at a constant rate.
The primary variable in free fall is gravitational acceleration, denoted as "g," which on the surface of the Earth has a value of approximately 9.8 meters per second squared (m/s虏). During free fall, an object continuously speeds up, starting from an initial velocity of zero if dropped at rest, as seen in the exercise.
  • Objects in free fall feel as if they are weightless because gravity is the only force acting on them.
  • The velocity of an object in free fall increases by approximately 9.8 m/s every second.
  • All objects, regardless of their mass, fall at the same rate in a vacuum.
Kinematic Equations
Kinematic equations describe the motion of objects and are particularly useful in solving problems related to linear motion with constant acceleration. These equations allow us to calculate unknown variables such as time, displacement, velocity, and acceleration. In the case of the falling bolt, we use them to determine how long it takes to fall certain distances and its velocity at various points.
Some important kinematic equations include:
  • The equation for displacement: \( s = ut + \frac{1}{2} a t^2 \) describes how far an object moves. In our exercise, it helps find the time to fall the last 20 meters.
  • The equation for final velocity: \( v^2 = u^2 + 2as \) gives the speed of the object after falling a certain distance.
  • The basic equation for velocity: \( v = u + at \) calculates how fast an object is moving after a specific time.
Remember, "s" stands for displacement, "u" is initial velocity, "v" is final velocity, "a" is acceleration, and "t" represents time. By substituting the known values into these equations, we find everything needed for the falling bolt problem.
Gravitational Acceleration
Gravitational acceleration is the acceleration of an object caused by Earth's gravity. It is constant and equal to 9.8 m/s虏 near the Earth's surface. This consistent force is why objects speed up at the same rate regardless of their mass鈥攚hether it is a heavy bolt or a rose petal. Gravitational acceleration affects all moving objects by increasing their velocity in a downward direction. We see this in the exercise where the bolt's speed increases steadily during its fall.
Gravitational acceleration is crucial when calculating:
  • How long it takes for objects to fall a certain distance.
  • The speed at which an object is moving at a particular moment.
  • The impact speed when the object reaches the ground.
Knowing this value allows us to use kinematic equations to find other unknowns and solve complex motion problems in a relatively straightforward way. Understanding this concept is key to mastering topics in both physics and everyday life observations, such as why an apple falls from a tree in the way it does.

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Most popular questions from this chapter

A particle confined to motion along an \(x\) axis moves with constant acceleration from \(x=2.0 \mathrm{~m}\) to \(x=8.0 \mathrm{~m}\) during a \(2.5 \mathrm{~s}\) time interval. The velocity of the particle at \(x=8.0 \mathrm{~m}\) is \(2.8 \mathrm{~m} / \mathrm{s}\). What is the constant acceleration during this time interval?

A world's land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at \(1020 \mathrm{~km} / \mathrm{h}\). He and the sled were brought to a stop in \(1.4 \mathrm{~s}\). (See Fig. 2-7.) In terms of \(g\), what acceleration did he experience while stopping?

When a high-speed passenger train traveling at \(161 \mathrm{~km} / \mathrm{h}\) rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance \(D=676 \mathrm{~m}\) ahead (Fig. 2-20). The locomotive is moving at \(29.0 \mathrm{~km} / \mathrm{h}\). The engineer of the high-speed train immediately applies the brakes. (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at \(x=0\) when, at \(t=0\), he first spots the locomotive. Sketch \(x(t)\) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.

In a particle accelerator, an electron enters a region in which it accelerates uniformly in a straight line from a speed of \(4.00 \times 10^{5}\) \(\mathrm{m} / \mathrm{s}\) to a speed of \(6.00 \times 10^{7} \mathrm{~m} / \mathrm{s}\) in a distance of \(3.00 \mathrm{~cm}\). For what time interval does the electron accelerate?

Two particles move along an \(x\) axis. The position of particle 1 is given by \(x=6.00 t^{2}+\) \(3.00 t+2.00\) (in meters and seconds); the acceleration of parti- \(\quad\) cle 2 is given by \(a=-8.00 t\) (in meters per second squared and seconds) and, at \(t=0\), its velocity is \(15 \mathrm{~m} / \mathrm{s}\). When the velocities of the particles match, what is their velocity?
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