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Chapter 5: Problem 19

A \(500 \mathrm{~kg}\) 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?

Short Answer

Expert verified
The magnitude of the required net force is 123455 N.

Step by step solution

01

Convert final velocity to m/s

The final velocity is given as \(1600\, \text{km/h}\). To convert this to meters per second, use the conversion where \(1 \text{ km/h} = \frac{1}{3.6}\text{ m/s}\). Thus, \(1600 \times \frac{1}{3.6} = 444.44\, \text{m/s}\).
02

Find acceleration

The rocket sled starts from rest, so the initial velocity \(u = 0\). Using the formula for acceleration \(a = \frac{v - u}{t}\), where \(v\) is the final velocity (\(444.44\, \text{m/s}\)) and \(t = 1.8\, \text{s}\), substitute the values: \(a = \frac{444.44 - 0}{1.8} = 246.91\, \text{m/s}^2\).
03

Calculate net force

Using Newton's second law \(F = ma\), where \(m = 500\, \text{kg}\) and \(a = 246.91\, \text{m/s}^2\), substitute the values: \(F = 500 \times 246.91 = 123455\, \text{N}\).

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Key Concepts

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

Conversion of Units
To solve physics problems, it's vital to have all measurements in compatible units. This often involves converting from one unit to another. In the rocket sled example, the speed is given in kilometers per hour (km/h), but we need it in meters per second (m/s) for further calculations. Unit conversion follows straightforward steps:
  • Know the conversion factor. For speed, 1 km/h equals \(\frac{1}{3.6}\) m/s.
  • Multiply the original value by the conversion factor.
Applying this, the sled's speed of 1600 km/h converts to meters per second by multiplying with \(\frac{1}{3.6}\), resulting in \(1600 \times \frac{1}{3.6} = 444.44\) m/s. This allows us to accurately compute subsequent steps in consistent units.
Acceleration
Acceleration is the rate at which an object changes its velocity. It is crucial in understanding motion dynamics. In this case, we calculate acceleration for a rocket sled that accelerates from rest.The formula to find acceleration is: \[ a = \frac{v - u}{t} \] where
  • \(v\) is the final velocity, \(444.44\, \text{m/s}\)
  • \(u\) is the initial velocity, here \(0\, \text{m/s}\) because it starts from rest
  • \(t\) is the time taken, which is \(1.8\, \text{s}\)
Plugging these values into the formula gives: \[ a = \frac{444.44 - 0}{1.8} = 246.91\, \text{m/s}^2 \] This solution gives us the constant acceleration the sled experiences over the specified time period.
Net Force Calculation
Newton's Second Law of Motion provides a direct link between force, mass, and acceleration. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:\[ F = ma \] This equation allows us to find the necessary force to achieve a certain acceleration. In our problem:
  • Mass \(m\) of the rocket sled is \(500\, \text{kg}\).
  • Acceleration \(a\) from our previous calculation is \(246.91\, \text{m/s}^2\).
Substituting these values, we find the net force \(F\):\[ F = 500 \times 246.91 = 123455 \text{ N} \]This force value signifies the magnitude of thrust needed to propel the rocket sled to the desired speed within the given time frame. Understanding this calculation helps illustrate how Newton's laws govern motion in dynamic systems.

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Most popular questions from this chapter

A block of mass \(m_{1}=3.70\) \(\mathrm{kg}\) on a frictionless plane inclined at angle \(\theta=30.0^{\circ}\) is connected by \(\varepsilon\) cord over a massless, frictionless pulley to a second block of mass \(m_{2}=2,30 \mathrm{~kg}\) (Fig. \(5-52\) ). What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?

Figure \(5-47\) shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time \(t=0\), container 1 has mass \(1.30 \mathrm{~kg}\) and container 2 has mass \(2.80 \mathrm{~kg}\), but container 1 is losing mass (through a leak) at the constant rate of \(0.200 \mathrm{~kg} / \mathrm{s}\). At what rate is the acceleration magnitude of the containers changing at (a) \(t=0\) and (b) \(t=3.00 \mathrm{~s}\) ? (c) When does the acceleration reach its maximum value?

A hot-air balloon of mass \(M\) is descending vertically with downward acceleration of magnitude \(a\). How much mass (ballast) must be thrown out to give the balloon an upward acceleration of magnitude \(a\) ? Assume that the upward force from the air (the lift) does not change because of the decrease in mass.

Only two horizontal forces act on a \(3.0 \mathrm{~kg}\) body that can move over a frictionless floor. One force is \(9.0 \mathrm{~N}\), acting due east, and the other is \(8.0 \mathrm{~N}\), acting \(62^{\circ}\) north of west. What is the magnitude of the body's acceleration?

Holding on to a towrope moving parallel to a frictionless ski slope, a \(50 \mathrm{~kg}\) skier is pulled up the slope, which is at an angle of \(8.0^{\circ}\) with the horizontal. What is the magnitude \(F_{\text {ropo }}\) of the force on the skier from the rope when (a) the magnitude \(v\) of the skier's velocity is constant at \(2.0 \mathrm{~m} / \mathrm{s}\) and (b) \(v=2.0 \mathrm{~m} / \mathrm{s}\) as \(v\) increases at a rate of \(0.10 \mathrm{~m} / \mathrm{s}^{2} ?\)
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