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hammer [34]
2 years ago
7

1. Ignore friction and determine an expression for the distance d the boxes travel before coming to rest. Choose the floor as th

e position for zero gravitational potential energy. (Use the following as necessary: m1, m2, v, and g for the acceleration due to gravity.)
2. Determine an expression for the work done on box 1 by the rope. (Use the following as necessary: m1, m2, v, and g for the acceleration due to gravity.)

Physics
1 answer:
Nadusha1986 [10]2 years ago
3 0

Newton's second law allows us to find the results for the displacement and work on box 1 are:

       1) The displacement is   x = \frac{v_o^2 m_2}{2(m_1+m_2)} g

      2) The work is  W= \frac{1}{2} m_1v_o^2

1) Newton's second law says that force is directly proportional to the mass and acceleration of bodies.

         F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the bodies

The reference system is a coordinate system with respect to which the decompositions of the forces are carried out and all the measurements are made, in this case we take a system where the x axis is horizontal, the y axis is vertical and the zero of the system is located at soil

In the attached we can see a free body diagram of the system where the external forces are indicated, let's apply Newton's second law to body 1

body 1

x-axis

          -T = m₁ a

body 2

 y-axis

           T - W₂ = m₂ a

           W₂ = m₂ g

let's solve the system

        - m₂ g = (m₁ + m₂) a

           a = - \frac{m_2}{m_1+m_2} \ g- m2 / m1 + m2 g

Kinematics studies the movement of bodies, let's use the expression

            v² = v_o^2 - 2 a x

When the body stops the velocity is zero

            x = \frac{v_o^2}{2a}

We substitute

            x = \frac{v_o^2}{2} \frac{m_2}{m_1+m_2} \ g

Since the two boxes are connected by a rope, they both travel the same distance

           

2) They ask for Tension work on box 1

Work is defined by the scalar product of force and displacement

        W = F. d

Where the bold letters indicate vectors, W is the work that is a scalar, F the force and d the displacement

We can write this expression by developing the dot product

       W  = F d cos θ

Where θ is the angle between force and displacement.

In this case the force is the tension of the rope, from the attached graph we see that the force is directed to the right and the displacement is to the left, therefore the angle is 180º and the cos  180 is equal to -1

We look for the tension from Newton's second law for box 1

          T = - m₁ a

Let's substitute

          T =- m_1 \ ( - \frac{m_2}{m_1+m_2} ) \ g  

          T = \frac{m_1m_2}{m_1+m_2} \ g

         

We calculate the work

         W = - \frac{m_1m_2}{m_1+m_2 } \ g ( \frac{v_o^2 (m1+m_2)}{2 m_2 g} )

         W = - ½ m₁ v₀²

In conclusion using Newton's second law we can find the results for the displacement and work on box 1 are:

1)  The displacement is  x= \frac{v_o^2 m_2}{2(m_1+m_2)} g

2) The work is  W = - ½ m₁ v₀²

Learn more here: brainly.com/question/17290735

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What happens to the iron in the coilgun if the electricity in the coil was turned on
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  • <em><u>The piece of iron has become a magnet. Some substances can be magnetized by an electric current. When electricity runs through a coil of wire, it produces a magnetic field. The field around the coil will disappear, however, as soon as the electric current is turned off.</u></em>
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2 years ago
A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55
SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → W=2,55.0,71=1,8105\frac{N}{m}

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = \frac{1}{2} mV^{2} + mgh

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

In our situation

\frac{0,71}{sin90} =\frac{h}{sin32} → h=0,376

Therefore

1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376

→ V=2,416

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3 years ago
Cells &amp; Cellular Processes
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Answer:

one or more cells

Explanation:

I need to have 20 characters to submit lol

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3 years ago
A 13,000-N vehicle is to be lifted by a 25-cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter pis
MaRussiya [10]

Answer:

Explanation:

We shall apply Pascal's Law in fluid mechanics

According to it , pressure is transmitted in liquid from one point to another without any change .

25 cm diameter = 12.5 x 10⁻² m radius

Area = 3.14 x (12.5 x 10⁻²)²

= 490.625 x 10⁻⁴ m²

Pressure by vehicle

Force / area

13000 / 490.625 x 10⁻⁴

= 26.497 x 10⁴ Pa

5 cm diameter = 2.5 x 10⁻² radius

area = 3.14 x (2.5 x 10⁻²)²

= 19.625 x 10⁻⁴ m²

If we assume required force F on this area

Pressure = F / 19.625 x 10⁻⁴ Pa

According to Pascal Law

F / 19.625 x 10⁻⁴  = 26.497 x 10⁴

F = 19.625 x 26.497

= 520 N

4 0
3 years ago
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