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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 required net force is 123455 N.

Step by step solution

01

Convert units

The final velocity is given in kilometers per hour (km/h), which needs to be converted to meters per second (m/s) because we are working within the SI unit system. Use the conversion factor: 1 km/h = (1/3.6) m/s. Convert 1600 km/h to m/s: 1600 km/h = 1600 x (1/3.6) m/s = 444.44 m/s.
02

Determine acceleration

Since the rocket sled starts from rest, its initial velocity (\(v_i\) is 0 m/s. The final velocity (\(v_f\)) is 444.44 m/s as calculated.We use the formula for acceleration \(a\) when the time \(t\) and initial and final velocities are known:\[a = \frac{v_f - v_i}{t}\]Substitute the known values: \ \(a = \frac{444.44 \, \text{m/s} - 0 \, \text{m/s}}{1.8 \, s} = 246.91 \, \text{m/s}^2\).
03

Calculate net force

Use Newton鈥檚 second law to find the net force (\(F\)) required to accelerate the sled. The mass \(m\) of the sled is 500 kg, and the acceleration \(a\) is 246.91 m/s虏.\[F = m \, a\]Substitute the known values: \ \(F = 500 \, \text{kg} \, \times \, 246.91 \, \text{m/s}^2 = 123455 \, \text{N}\).

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Key Concepts

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

Velocity Conversion
Understanding unit conversion is essential, especially when dealing with velocity in different measurement systems. The original exercise provided the final velocity of the rocket sled in kilometers per hour (km/h), but calculations within the International System of Units (SI) require meters per second (m/s). This is because the standard unit of distance in SI is meters, and time is seconds.

To convert velocity from kilometers per hour to meters per second, use the conversion factor **1 km/h = \(\frac{1}{3.6}\) m/s**. This factor arises from calculations knowing that 1 kilometer equals 1000 meters and 1 hour equals 3600 seconds. Therefore, to convert 1600 km/h to m/s, you multiply by the conversion factor:
  • Start with 1600 km/h.
  • Multiply by \(\frac{1}{3.6}\).
  • Resulting in 444.44 m/s.
This step ensures that all the units in our calculations match, keeping the arithmetic simple and the results consistent.
Acceleration Calculation
Acceleration measures the rate of change of velocity over time. Since the exercise states the rocket sled starts from rest, meaning its initial velocity (Let鈥檚 break down the formula used in the solution: **\[ a = \frac{v_f - v_i}{t} \]** , where:
  • \(v_f\) is the final velocity - 444.44 m/s as converted.
  • \(v_i\) is the initial velocity - 0 m/s as the sled starts from rest.
  • \(t\) is the time over which acceleration occurs - 1.8 seconds in this case.
By plugging these numbers into the formula, we calculate the sled's acceleration as **246.91 m/s虏**. This value means that each second, the sled's velocity increases by 246.91 meters per second. Understanding how each variable affects the acceleration helps in comprehending the dynamics involved in motion, especially in physics disciplines related to forces and dynamics.
Force Calculation
The final goal in the exercise is to determine the net force required to accelerate the rocket sled. This is where Newton鈥檚 Second Law of Motion comes into play, which is succinctly expressed as **\[ F = ma \]**, meaning force equals mass times acceleration.

Here's how it breaks down in the solution:
  • The sled鈥檚 mass \(m\) is given - 500 kg.
  • Acceleration \(a\) has been calculated - 246.91 m/s虏.
  • These values are substituted into the formula to find force \(F\).
The calculation \(F = 500 \, \text{kg} \times 246.91 \, \text{m/s}^2 = 123455 \, \text{N}\) gives us a net force of 123,455 Newtons. This means that in order for the sled to reach the desired velocity within the given time frame, a force of this magnitude must be applied. Understanding this concept is crucial in engineering and physical science, where controlling object motion is essential. It illustrates the scalability of force in relation to mass and acceleration.

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Most popular questions from this chapter

Using a rope that will snap if the tension in it exceeds \(387 \mathrm{~N}\), you need to lower a bundle of old roofing material weighing \(449 \mathrm{~N}\) from a point \(6.1 \mathrm{~m}\) above the ground. Obviously if you hang the bundle on the rope, it will snap. So, you allow the bundle to accelerate downward. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

The velocity of a \(3.00 \mathrm{~kg}\) par.ticle is given by \(\vec{v}=\left(8.00 t \hat{\mathrm{i}}+3.00 t^{2} \hat{\mathrm{j}}\right)\) \(\mathrm{m} / \mathrm{s},\) with time \(t\) in seconds. At the instant the net force on the particle has a magnitude of \(35.0 \mathrm{~N},\) what are the direction (relative to the positive direction of the \(x\) axis) of (a) the net force and (b) the particle's direction of travel?

A \(40 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal. Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{x}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2},\) and (c) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

Figure \(5-45\) shows a \(5.00 \mathrm{~kg}\) block being pulled along a frictionless floor by a cord that applies a force of constant magnitude \(20.0 \mathrm{~N}\) but with an angle \(\theta(t)\) that varies with time. When angle \(\theta=25.0^{\circ},\) at what rate is the acceleration of the block changing if (a) \(\theta(t)=\left(2.00 \times 10^{-2} \mathrm{deg} / \mathrm{s}\right) t \quad\) and (b) \(\theta(t)=-\left(2.00 \times 10^{-2} \mathrm{deg} / \mathrm{s}\right) t ?\) (Hint: The angle should be in radians.)

An elevator cab and its load have a combined mass of \(1600 \mathrm{~kg}\). Find the tension in the supporting cable when the cab, originally moving downward at \(12 \mathrm{~m} / \mathrm{s}\), is brought to rest with constant acceleration in a distance of \(42 \mathrm{~m}\).
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