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Chapter 4: Problem 17

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes \(28 s\) to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular?

Short Answer

Expert verified
The probe takes 19.8 seconds to travel the same distance with perpendicular forces.

Step by step solution

01

Understand the Physics

The problem involves understanding how force and acceleration work in straight-line and perpendicular directions. When two forces act in the same direction, they sum up to provide maximum acceleration. When they act perpendicularly, they behave independently, affecting motion components separately.
02

Apply Newton's Second Law for Same Direction Case

When the engines are fired in the same direction, their forces add up. Let each engine apply a force of \( F \), so together they apply a force \( 2F \). According to Newton's second law, \( F = ma \), where \( m \) is the mass and \( a \) is the acceleration. Therefore, \( 2F = ma_1 \).
03

Calculate the Required Time for Same Direction Force

Using the kinematic equation for distance \( s = \frac{1}{2}a_1t^2 \) where \( t = 28 \, s \), we can express it as \( s = \frac{1}{2}(\frac{2F}{m})(28)^2 \). This simplifies to \( s = \frac{56F}{m} \times 28 \).
04

Apply Newton's Laws for Perpendicular Forces

When the forces are perpendicular, each force \( F \) causes motion in its direction, leading to two independent accelerations, \( a_x = \frac{F}{m} \) and \( a_y = \frac{F}{m} \). The resultant acceleration is \( a_r = \sqrt{a_x^2 + a_y^2} = \sqrt{2} \frac{F}{m} \).
05

Calculate the Time for Perpendicular Forces

Using the kinematic equation \( s = \frac{1}{2}a_r t^2 \) where \( a_r = \sqrt{2} \frac{F}{m} \), we need to solve for \( t \). Substitute the expression for distance: \( s = \frac{56F}{m} \) which gives \( \frac{56F}{m} = \frac{1}{2}\sqrt{2} \frac{F}{m} t^2 \). Solve for \( t \) to find \( t = \frac{28}{\sqrt{2}} = 19.8 \, s \).
06

Conclusion

With the engines at perpendicular directions, the resultant time is \( 19.8 \, s \) to travel the same distance.

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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 classical mechanics that describes the motion of objects without considering the causes of motion. It deals with quantities such as position, velocity, and acceleration. These concepts help us understand how an object's motion changes over time. In this exercise, we use kinematics to determine the time it takes for a space probe to travel a certain distance under different force conditions. By applying kinematic equations, such as
  • \(s = \frac{1}{2} a t^2\), where \(s\) is the distance, \(a\) is the acceleration, and \(t\) is time
we can analyze how the probe's motion is affected when subjected to different force arrangements.
Perpendicular Forces
Perpendicular forces occur when two forces act at 90 degrees to each other. In physics, particularly in Newtonian mechanics, each force affects motion independently in its respective direction. Consider a situation where a space probe's engines exert perpendicular forces. Each engine creates acceleration along its line of action:
  • Let each force be \(F\), resulting in accelerations \(a_x = \frac{F}{m}\) in the x-direction and \(a_y = \frac{F}{m}\) in the y-direction, where \(m\) is the mass.
These two independent accelerations mean the probe's movement is not straightforward, requiring us to calculate the resultant effect on the motion.
Resultant Acceleration
Resultant acceleration is the combined effect of multiple individual accelerations acting on a body. For perpendicular forces, it is determined using the Pythagorean theorem due to the orthogonal nature of the forces involved. In our example, the resultant acceleration \(a_r\) can be calculated by:
  • \(a_r = \sqrt{a_x^2 + a_y^2} = \sqrt{2}\frac{F}{m}\).
This gives us a single value that represents how fast the probe will accelerate as a result of the two perpendicular forces. Having \(a_r\), we can then plug it into the kinematic equation to find the time it takes to travel the same distance as when forces are aligned.

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Most popular questions from this chapter

A raindrop has a mass of \(5.2 \times 10^{-7} \mathrm{kg}\) and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.

ssm Two forces \(\overrightarrow{\mathbf{F}}_{\mathrm{A}}\) and \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\) are applied to an object whose mass is 8.0 \(\mathrm{kg}\) . The larger force is \(\overrightarrow{\mathbf{F}}_{\mathrm{A}}\) . When both forces point due east, the object's acceleration has a magnitude of 0.50 \(\mathrm{m} / \mathrm{s}^{2}\) . However, when \(\overline{\mathrm{F}}_{\mathrm{A}}\) points due east and \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\) points due west, the acceleration is 0.40 \(\mathrm{m} / \mathrm{s}^{2}\) due east. Find (a) the magnitude of \(\overrightarrow{\mathbf{F}}_{\mathbf{A}}\) and \(\quad(\mathbf{b})\) the magnitude of \(\overrightarrow{\mathbf{F}}_{\mathbf{B}}\).

On earth, two parts of a space probe weigh 11000 \(\mathrm{N}\) and 3400 \(\mathrm{N}\) . These parts are separated by a center-to-center distance of 12 \(\mathrm{m}\) and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

A \(5.00-\mathrm{kg}\) block is placed on top of a \(12.0-\mathrm{kg}\) block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.600 . What is the maximum horizontal force that can be applied before the \(5.00-\mathrm{kg}\) block begins to slip relative to the 12.0 \(\mathrm{kg}\) block, if the force is applied to \((\text { a the more massive block and }\) less massive block?

The principles used to solve this problem are similar to those in Multiple- Concept Example 17. A 205-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0 with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.900, and the log has an acceleration of magnitude 0.800 \(\mathrm{m} / \mathrm{s}^{2}\) . Find the tension in the rope.
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