Physics, 02.07.2019 08:30, lesheaux

Asolid block, with a mass of 8.58kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring constant value of 368n/m, until the spring is compressed by 11cm. the block is then released. what is the block's final speed, in m/s?

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In this problem, we are talking about Mechanical Energy () which is the addition of the Kinetic Energy (energy of the body in motion) and Potential Energy (It can be Gravitational Potential Energy or Elastic Potential Energy, in this case is the second):

M=K+P (1)

The Kinetic Energy is: $K=\frac{1}{2}mV^{2}$

Where is the mass of the body and its velocity

And the Potential Energy (Elastic) is: $P=\frac{1}{2}cX^{2}$

Where is the spring constant and is the the position of the body.

Knowing this, the equation for the Mechanical Energy in this case is:

$M=\frac{1}{2}mV^{2}+\frac{1}{2}cX^{2}$ (2)

Now, according to the Conservation of the Energy Principle, and knowing there is not friction, the initial energy $M_{i}$ must be equal to the final energy $M_{f}$ :

$M_{i}=M_{f}$ (3)

$M_{i}=\frac{1}{2}m{V_{i}}^{2}+\frac{1}{2}c{X_{i}^{2}}$ (4)

At the beginning, the block has a $V_{i}=0$ , because it starts from rest, this means the initial energy $M_{i}$ is only the Potential Elastic Energy:

$M_{i}=\frac{1}{2}c{X_{i}^{2}}$ (5)

After the spring is compressed, is in its equilibrium point X=0, so $X_{f}=0$ . Then the block is released. This means the final energy $M_{f}$ is only the Kinetic Energy

$M_{f}=\frac{1}{2}m{V_{f}}^{2}$ (6)

Now, we have to substitute (5) and (6) in (3):

$\frac{1}{2}c{X_{i}^{2}}=\frac{1}{2}m{V_{f}}^{2}$

$V_{f}=\sqrt{\frac{c{X_{i}^{2}}}{m}}$

$V_{f}= X_{i}\sqrt{\frac{c}{m}}$

$V_{f}=0.11m\sqrt{\frac{368N/m}{8.58kg}}$

Finally:

$V_{f}=0.7203m/s$

answered: Guest

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

answered: Guest

Answertraveling around a circular trackexplanation"traveling around a circular track" is the best answer as you could be traveling around a circular track at a constant speed. your velocity would therefore have the same magnitude but will always be constantly direction. since acceleration is the rate of change of velocity, the rate of change will only involve the direction of velocity and not the magnitude.wrong choices: "an object at rest" does not accelerate at all"increasing speed while traveling around a curve" means that the acceleration will not be constant due to only change in direction; it will be constant due to both magnitude and direction"an object in free fall" does not necessarily change direction

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Asolid block, with a mass of 8.58kg, on a frictionless surface is pushed directly onto a horizontal...