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torisob [31]
3 years ago
13

A 2.85 kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 875 N/m. The spring

initially is neither stretched nor compressed. The fish is released from rest.A)What is its speed after it has descended 0.0580 m from its initial position? B)What is the maximum speed of the fish as it descends?
Physics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

A) V = 0.324 m/s

B) V = 0.56 m/s

Explanation:

A) Change in gravitational potential energy = mass * gravity * height change

Change in G.P.E = 2.85 * 9.81 * 0.058 = 1.6215 J

Energy transferred to spring = 0.5 * k * x^2

Here k = 875 N/m

And x = 0.058 m

Thus energy transferred to spring = 1.47175 J

The G.P.E is converted to spring potential energy and kinetic energy as it moves down. So the difference in energy is accounted by kinetic energy as follows:

Kinetic energy = G.P.E - Spring P.E

Kinetic energy = 1.6215 - 1.47175 = 0.14975 J

We can now find the speed using this kinetic energy:

Kinetic energy = 0.14975

0.5 * mass * velocity^2 = 0.14975

0.5 * 2.85 * V^2 = 0.14975

V = 0.324 m/s

B) The fish accelerates because the force on it are unbalanced. These forces are the weight of the fish, and the force of the spring stopping it.

As long as the weight of the fish is more than the upward force of the spring, the fish will continue to accelerate. Using this knowledge, we can deduce that the speed is maximum when the weight and spring force are equal. Thus we set them equal and find out the displacement first:

Weight = spring force

2.85 * 9.81 = 875 * Displacement

Displacement = 0.03195 m

Similarly as (A):

Change in G.P.E = 2.85 * 9.81 * 0.03195 = 0.8933 J

Spring P.E = 0.5 * 875 * 0.03195^2 = 0.4466 J

Kinetic energy = 0.8933 - 0.4466 = 0.4467 J

0.5 * mass * V^2 = 0.4467

0.5 * 2.85 * V^2 = 0.4467

V = 0.56 m/s

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A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a
grigory [225]

Answer:

a)  fr = 266.92 N,   fy = 1300 N,  b)    μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

             -w x - W x₂ + R y = 0         (1)

usemso trigonometry to find distances

            cos 60.08 = x / 7.5

            x = 7.5 cos 60.08

            x = 3.74 m

fireman

           cos 60.08 = x₂ / 4

           x2 = 4 cos 60

           x2 = 2 m

wall support

           sin 60.08 = y / 15

           y = 15 are 60.08

           y = 13 m

we substitute in equation 1

           R y = w x + W x2

            R = (w x + W x2) / y

            R = (500 3.74 +800 2) / 13

            R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

           R -fr = 0

           R = fr

           fr = 266.92 N

Y Axis  

           Fy - w-W = 0

           fy = 500 + 800

           fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

          cos 60.08 = x₃ / 9.00

          x₃ = 9 cos 60

          x₃ = 5.28 m

from equation 1

          R = (w x + W x₃) / y

          R = 500 3.74 + 800 5.28) / 13

          R = 468.769 N

we saw that

          fr = R = 468.769

The expression for the friction force is

          fr = μ N

in this case the normal is the ratio to pesos

        N = Fy

       N = 1300 N

        μ N = fr

        μ = fr / N

        μ = 468,769 / 1300

         μ = 0.36

7 0
3 years ago
2. A person lifts 200kg seven times over the course of 11.8s. If they displaced the weight 2.2m up each time, how much power did
Aneli [31]

Answer:

<em>The person delivered a power of 2,558 Watt</em>

Explanation:

<u>Work and Power</u>

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being  the force vector and  the displacement vector, the work is calculated as:

W=\vec F\cdot \vec s

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy transferred per unit of time. In the SI, the unit of power is the watt, equivalent to one joule per second.

The power can be calculated as:

\displaystyle P=\frac {W}{t}

Where W is the work and t is the time.

If the person lifts a mass of m=200 Kg, then exerts a force equal to its weight:

F = m.g = 200*9.8 = 1,960

F = 1,960 N

The work done when lifting the weight 7 times by a distance of s=2.2 m is:

W = 7*1,960*2.2=30,184

W = 30,184 J

Finally, the power delivered in t=11.8 seconds is:

\displaystyle P=\frac {30,184}{11.8}

P = 2,558 Watt

The person delivered a power of 2,558 Watt

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Explanation:

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<span>Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance. We calculate as follows:

</span><span>Volume = 60.0 g ( 1 mL / 0.70 g ) = 85.71 mL

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