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Licemer1 [7]
3 years ago
7

An object with mass M is attached to the

Physics
2 answers:
zysi [14]3 years ago
8 0

Answer:

W = 5/4 Mgℓ

Explanation:

Sum of forces:

∑F = ma

T − Mg = M (g/4)

T = 5/4 Mg

Work = force × distance

W = Tℓ

W = 5/4 Mgℓ

olchik [2.2K]3 years ago
8 0

Work done by the tension is ⁵/₄ MgL

\texttt{ }

<h3>Further explanation</h3>

Let's recall the Kinetic Energy and Work formula as follows:

\boxed {E_k = \frac{1}{2}mv^2 }

<em>Ek = kinetic energy ( J )</em>

<em>m = mass of object ( kg )</em>

<em>v = speed of object ( m/s )</em>

\texttt{ }

\boxed {W = F d}

<em>W = work done ( J )</em>

<em>F = force ( N )</em>

<em>d = displacement ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

acceleration of the object = a = g/4

initial speed of the object = u = 0 m/s

distance traveled = d = L

<u>Asked:</u>

work done by the tension = W = ?

<u>Solution:</u>

We will use Newton's Law of Motion to solve this problem:

\Sigma F = Ma

T - w = Ma

T = w + Ma

T = Mg + M\frac{g}{4}

T = \frac{5}{4}Mg

T L = \frac{5}{4} MgL

W = \frac{5}{4} MgL

\texttt{ }

<h3>Conclusion :</h3>

Work done by the tension is ⁵/₄ MgL

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Energy

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Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

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Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

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\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

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\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

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\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

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