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Anna11 [10]
3 years ago
6

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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A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.
irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system (\tau), measured in Newton-meters, is:

\tau = I\cdot \alpha (1)

Where:

I - Moment of inertia, measured in Newton-meter-square seconds.

\alpha - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

\alpha = \frac{\omega - \omega_{o}}{t} (2)

Where:

\omega_{o} - Initial angular speed, measured in radians per second.

\omega - Final angular speed, measured in radians per second.

t - Time, measured in seconds.

If we know that \tau = 3\,N\cdot m, \omega_{o} = 0\,\frac{rad}{s }, \omega = 145.875\,\frac{rad}{s} and t = 4\,s, then the moment of inertia of the motor is:

\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}

\alpha = 36.469\,\frac{rad}{s^{2}}

I = \frac{\tau}{\alpha}

I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }

I = 0.0823\,N\cdot m\cdot s^{2}

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0
3 years ago
What is a process that returns to its beginning and repeat it self in the same sequence
jasenka [17]

Answer:

a sequence

Explanation: was that an option?, or was it not a multiple choice question ?

7 0
3 years ago
For the following inclined plane, m1 = 5.0 kg, m2 = 6.0 kg, and a = 30° A. Draw force diagrams for blocks m1 and m2. B. Calculat
sattari [20]

Answer:

See attached

Explanation:

6 0
2 years ago
A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?
iris [78.8K]

Answer:

\Delta x=22.67786838m

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2 (1)

Where:

y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis

y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis

v_o=initial\hspace{3}velocity

t=travel\hspace{3}time

g=gravity\hspace{3}constant

\theta=Initial\hspace{3}launch\hspace{3}angle

In this case:

\theta=0

Because the dog jumps horizontally

Let's asume the gravity constant as:

g=9.8

y=0

Because when the dog reach the base the height is 0

y_o=70

v_o=6

Now let's replace the data in (1)

y_o-\frac{1}{2} *(9.8)*t^2+70

Isolating t:

t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473

Finally let's find the horizontal displacement using this equation:

\Delta x=v_o*cos(\theta)*t

Replacing the data:

\Delta x=6*1*3.77964473=22.67786838m

8 0
3 years ago
PLS Can someone help me with this I don’t know how to do it. I’ll give brainliest to whoever answers it.
Stolb23 [73]

Answer:

I think this is how you do it

Explanation:

4 0
3 years ago
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