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2 years ago

Calculate the force applied to raise a table to a height of 8m if 120j of work was done

1 answer:

7 0

The force applied to raise a table to a height of 8m when 120 Joules of work is done is 15 N. That is, Force F = 15N

Work is the product of force and distance in the direction of force. Work is done on an object when the displacement of the object is in the direction of the force applied. That is,

W = F x S

The S.I unit of work is Joule.

Given that a table to a height of 8m and 120J of work was done. To find the force applies, the parameters to consider are

Work W = 120 J

Distance S = 8 m

Force F = ?

Make force F the subject of formula from the formula written above. That is,

F = W/S

F = 120 / 8

F = 15 N

Therefore, if the 120 Joule of work is done, the force applied to raise a table to a height of 8m is 15 Newtons.

Learn more about Work here: brainly.com/question/25573309

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How many cubic meters of habitable space was available in skylab?

loris [4]

361 cubic meters of habitable space was available in the skylab. None of the given options are correct.

<h3>Who launched skylab?</h3>

NASA launched the first American space station, called Skylab.

The fuel tank of the Saturn SIV-B rocket stage served as the Skylab's residential quarters.

Skylab's livable capacity was large in comparison to the spacecraft that American astronauts had previously used 361 cubic meters.

Skylab offered 361 cubic meters of livable area.

Hence, none of the given options is correct.

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4 0

2 years ago

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$