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Chapter 3: Problem 5

An experimental rocket sled can be accelerated at a constant rate from rest to \(1600 \mathrm{~km} / \mathrm{h}\) in \(1.8 \mathrm{~s}\). What is the magnitude of the required net force if the sled has a mass of \(500 \mathrm{~kg}\) ?

Short Answer

Expert verified
The required net force is approximately 123454.55 N.

Step by step solution

01

- Convert speed from km/h to m/s

First, convert the final speed from kilometers per hour to meters per second. We know that: 1 km/h = 1000/3600 m/s = 5/18 m/s Hence, 1600 km/h = 1600 * (5/18) m/s = 1600 * 5 / 18 = 444.44 m/s
02

- Identify the given values

Initial velocity (u_0) = 0 m/s (since it starts from rest) Final velocity (u) = 444.44 m/s Time (t) = 1.8 s Mass (m) = 500 kg
03

- Calculate the acceleration

Use the formula to calculate acceleration, a = (v - v_0) / ta = (444.44 m/s - 0 m/s) / 1.8 sa = 246.91 m/s^2
04

- Calculate the net force

Use Newton's second law, F = m * a F = 500 kg * 246.91 m/s^2 F = 123454.55 N

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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 describes the motion of objects without considering the forces that cause this motion. This involves parameters such as displacement, velocity, and acceleration.

In this problem, we deal with the rocket sled's initial and final velocities, as well as the time taken to reach the final velocity. These parameters fall under kinematics. Using these, we calculated the sled's acceleration.

Formulas used like \((v = u + at)\) and \((a = (v - u) / t)\) come from the equations of motion under constant acceleration. They allow us to determine how the sled's velocity changes over time.
Newton's second law
Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it is expressed as \((F = ma)\).

This law is pivotal in finding the net force required to accelerate the rocket sled. Given the mass \((m)\) and the calculated acceleration \((a)\), you directly apply Newton's second law to find the force. This concept connects the dots between the mass, the rate of change in velocity, and the force needed to induce that change.

In our example, this was illustrated by multiplying the mass of 500 kg by the acceleration of 246.91 m/s虏 to find the net force.
acceleration calculation
Acceleration is the rate of change of velocity of an object. It is calculated using the formula \((a = (v - u) / t)\), where \((v)\) is the final velocity, \((u)\) is the initial velocity, and \((t)\) is the time taken for this change.

In our problem, the sled starts from rest, implying the initial velocity \((u)\) is 0 m/s. The final velocity \((v)\) is given as 444.44 m/s, converted from 1600 km/h. The time \((t)\) is 1.8 seconds.

By substituting these values into the acceleration formula, we calculated the acceleration \((a)\), resulting in 246.91 m/s虏.
unit conversion
Unit conversion is a fundamental skill in physics, ensuring that all quantities are in compatible units for calculations.

In the given problem, the initial speed was provided in km/h. However, the standard SI unit for velocity used in calculations is m/s. The conversion factor used is 1 km/h = 5/18 m/s.

To convert the sled's speed from 1600 km/h to m/s, you multiply by 5/18:
\(1600 \times \frac{5}{18} = 444.44\) m/s.

Unit conversion ensures accurate calculations and avoids errors that can arise from mixing units. It is crucial in physics problems like this one where various parameters need to be consistent.

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Most popular questions from this chapter

Then Jumps A \(29.0 \mathrm{~kg}\) child, with a \(4.50 \mathrm{~kg}\) backpack on his back, first stands on a sidewalk and then jumps up into the air. Find the magnitude and direction of the force on the sidewalk from the child when the child is (a) standing still and (b) in the air. Now find the magnitude and direction of the net force on Earth due to the child when the child is (c) standing still and (d) in the air.

Newton's First Law states that an object will move with a constant velocity if nothing acts on it. This seems to contradict our everyday experience that a moving object comes to a rest unless something acts on it to keep it going. Does this everyday experience contradict Newton's First Law? If it does not, explain how this experience is consistent with Newton's First Law. If it does, explain why we bother to teach Newton's First Law anyway.

A ball is thrown vertically downward from the top of a \(36.6\) -m-tall building. The ball passes the top of a window that is \(12.2 \mathrm{~m}\) above the ground \(2.00 \mathrm{~s}\) after being thrown. What is the speed of the ball as it passes the top of the window?

Newton's Third Law says that objects that touch each other exert forces on each other. These forces satisfy the rule: If object \(A\) exerts a force on object \(B\), then object \(B\) exerts a force back on object \(A\) and the two forces are equal in magnitude but opposite in direction. Consider the following three situations concerning two identical cars and a much heavier truck. (a) One car is parked and the other car crashes into it. (b) One car is parked and the truck crashes into it. (c) The truck is pushing the car, because the car's cngine cannot start. The two are touching and the truck is speeding up. For each situation, do you think Newton's Third Law holds or does not hold? Explain your reasons for saying so.

A stone is dropped into a river from a bridge \(43.9 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. Both stones strike the water at the same time. (a) What is the initial speed of the second stone?
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