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3 years ago

What is the advantage in solving motion problems using energy conservation principles instead of free body diagrams

Physics

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Answer:

However, the disadvantages are:

1. Many atimes for some motion prolems, free-body diagrams has to be drawn many times so to have enough equations to solve for the unknowns. This is not the same with energy conservation principles.

2. In situations where we need to find the internal forces acting on an object, we can't truly solve such problems using free-body diagram as it captures external forces. This is not the same with energy conservation principles.

Explanation:

Often times the ideal method to use in solving motion problem related questions are mostly debated.

Energy conservation principles applies to isolated systems are useful when object changes their positions in moving upward or downward converts its potential energy due to gravity for kinetic energy, or the other way round. When energy in a system or motion remains constant that is energy is neither created nor destroyed, it can therefore be easier to calculate other unknown paramters like in the motion problem velocity, distance bearing it in mind that energy can only change from one type to another.

On the other hand, free body diagram which is a visual representation of all the forces acting on an object including their directions has so many advantages in solving motion related problems which include finding relationship between force and motion in identifying the force acting on a body.

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4 years ago

A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib

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The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

$K_i + U_i = K_f + U_f$

where

$K_i = 0$ is the kinetic energy of the tool at the top (zero since it is at rest)

$U_i = mgh$ is the gravitational potential energy of the tool at the top, where

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h is the heigth of the tool

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v is the final speed of the tool

$U_f = 0$ is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

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$mgh=\frac{1}{2}mv^2$

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And solving for v,

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3 years ago

What is the purpose of the zigzag line on the periodic table

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4 years ago

Read 2 more answers

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

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Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$ (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$

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3 years ago

A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa

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Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

$F-mg=ma$

Solving for the force F, we obtain that:

$F=ma+mg=m(a+g)$

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

$F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})$

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

$F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N$

In this case, the force from the rocket's thrusters is equal to 118,000N.

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4 years ago