Using Spring Equations - Physics

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Question

A spring with a spring constant of $200 \frac{N}{m}$ is compressed . What is the potential energy stored in the spring?

Answer

The equation for spring potential energy is $PE=\frac{1}{2}kx^2$ .

Plug in the given values for the distance and spring constant to solve for the potential energy.

$PE=(\frac{1}{2}) (200\frac{N}{m}) (- 0.5m)^2$

Remember, since the spring was compressed, it has a negative displacement. The resultant potential energy will be positive as, when released, the displacement will be along the positive horizontal axis.

$PE=(\frac{1}{2})(200\frac{N}{m}) (0.25m^2)$

$PE=\frac{1}{2}(50\frac{N\cdot m^2}{m})$

$PE=\frac{1}{2} (50Nm)$

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Question

A spring has a spring constant of $200\frac{N}{m}$ .

What force is required to compress it ?

Answer

For this problem, use Hooke's law:

$F=k\Delta x$

In this formula, is the spring constant, $\Delta x$ is the compression of the spring, and is the necessary force. We are given the values for the spring constant and the distance of compression. Using these terms, we can sovle for the force of the spring.

Plug in our given values and solve.

$F=k\Delta x$

$F=200\frac{N}{m}(-0.1m)$

Note that the force is negative because it is compressing the spring, pushing against the coil. When the force is released, the equal and opposite force of the spring will cause it to extend in the positive direction.

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Question

A spring has a spring constant of $16\frac{N}{m}$ .

If a force of is used to stretch out the spring, what is the total displacement of the spring?

Answer

For this problem, use Hooke's law:

$F=k\Delta x$

In this formula, is the spring constant, $\Delta x$ is the compression of the spring, and is the necessary force. We are given the spring constant and the force, allowing us to solve for the displacement.

Plug in our given values and solve.

$F=k\Delta x$

$20.5N=16\frac{N}{m}\Delta x$

$\frac{20.5N}{16\frac{N}{m}}=\Delta x$

$1.28m=\Delta x$

Note that both the force and the displacement are positive because the stretching force will pull in the positive direction. If the spring were compressed, the change in distance would have been negative.

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Question

A vertical spring with a spring constant of $11.1\frac{N}{m}$ is stationary. An mass is attached to the end of the spring. What is the maximum displacement that the spring will stretch?

$\small g=-9.8\frac{m}{s^2}$

Answer

The best way to solve this problem is by using energy. Notice that the spring on its own is stationary. That means its initial total energy at that moment is zero. When the mass is attached, the spring stretches out, giving it spring potential energy ( $PE_{spring}$ ).

Where does that energy come from? The only place it can come from is the addition of the mass. Since the system is vertical, this mass will have gravitational potential energy.

Use the law of conservation of energy to set these two energies equal to each other:

$PE_{mass}=PE_{spring}$

$mg\Delta y=\frac{1}{2}k\Delta y^2$

We are trying to solve for displacement, and now we have an equation in terms of our variable.

Start by diving both sides by $\Delta y$ to get rid of the $\Delta y^2$ on the right side of the equation.

$\frac{mg\Delta y}{\Delta y}=\frac{\frac{1}{2}k\Delta y^2}{\Delta y}$

$mg=\frac{1}{2}k\Delta y$

We are given values for the spring constant, the mass, and gravity. Using these values will allow use to solve for the displacement.

$mg=\frac{1}{2}k\Delta y$

$8kg* -9.8\frac{m}{s^2}=\frac{1}{2}* 11.1\frac{N}{m}* \Delta y$

$-78.4N=5.55\frac{N}{m}* \Delta y$

$\frac{-78.4N}{5.55\frac{N}{m}}=\Delta y$

$-14.13m=\Delta y$

Note that the displacement will be negative because the spring is stretched in the downward direction due to gravity.

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Question

How much force is required to compress a spring , if it has a spring constant of $200\frac{N}{m}$ ?

Answer

For this problem use Hooke's Law:

$F=k\Delta x$

Plug in our given values:

$F=200\frac{N}{m}*0.2m$

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Question

How much force is required to compress a spring , if it has a spring constant of $181\frac{N}{m}$ ?

Answer

For this problem use Hooke's Law:

$F=k\Delta x$

Plug in our given values:

$F=181\frac{N}{m}*9.32m$

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Question

How much potential energy is created by compressing a spring , if it has a spring constant of $200\frac{N}{m}$ ?

Answer

The formula for spring potential energy is:

$PE=\frac{1}{2}kx^2$

Plug in our given values and solve:

$PE=\frac{1}{2}200\frac{N}{m}(0.2m)^2$

$PE = \frac{1}{2}200\frac{N}{m}(0.04m^{2})$

$PE = \frac{1}{2}8Nm$

, so:

$PE=\frac{1}{2}*8J$

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Question

How much potential energy is generated by compressing a spring , if it has a spring constant of $181\frac{N}{m}$ ?

Answer

The formula for spring potential energy is

$PE=\frac{1}{2}kx^2$

Plug in our given values and solve:

$PE=\frac{1}{2}181\frac{N}{m}(9.32m)^2$

$PE=90.5\frac{N}{m}*86.8624m^2$

, so:

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Question

A mass is placed at the end of a spring. The spring is compressed . What is the maximum velocity of the mass if the spring has a spring constant of $200\frac{N}{m}$ ?

Answer

If we're looking for the maximum velocity, that will happen when all the energy in the system is kinetic energy.

We can use the law of conservation of energy to see . So, if we can find the initial potential energy, we can find the final kinetic energy, and use that to find the mass's final velocity.

The formula for spring potential energy is:

$PE=\frac{1}{2}kx^2$

Plug in our given values and solve:

$PE=\frac{1}{2}200\frac{N}{m}(0.2m)^2$

$PE=\frac{1}{2}200\frac{N}{m}(0.04m^{2})$

$PE=\frac{1}{2}8Nm$

, so:

$PE=\frac{1}{2}8J$

The formula for kinetic energy is:

$KE=\frac{1}{2}mv^2$

Since $PE_{i}=KE_{f}$ , that means that .

We can plug in that information to the formula for kinetic energy to solve for the maximum velocity:

$KE=\frac{1}{2}mv^2$

$4J=\frac{1}{2}0.56kgv^2$

$\frac{8J}{0.56kg}=v^2$

$14.29\frac{J}{kg} = v^{2}$

Since $J = \frac{kgm^{2}}{s^{2}}$ , that means that $\frac{J}{kg}=\frac{m^{2}}{s^{2}}$

$14.29\frac{m^2}{s^2}=v^2$

$\sqrt{14.29\frac{m^2}{s^2}}=\sqrt{v^2}$

$3.78\frac{m}{s}=v$

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Question

A mass is placed at the end of a spring. If the spring is compressed , what will be the mass's final velocity if the spring has a spring constant of $181\frac{N}{m}$ ?

Answer

If we're looking for the maximum velocity, that will happen when all the energy in the system is kinetic energy.

We can use the law of conservation of energy to see . So, if we can find the initial potential energy, we can find the final kinetic energy, and use that to find the mass's final velocity.

The formula for spring potential energy is:

$PE=\frac{1}{2}kx^2$

Plug in our given values and solve:

$PE=\frac{1}{2}181\frac{N}{m}(9.32m)^2$

$PE=90.5\frac{N}{m}86.8624m^2$

, so:

The formula for kinetic energy is:

$KE=\frac{1}{2}mv^2$ .

Since $PE_{i}=KE_{f}$ , we know that .

We can plug this information into the formula for kinetic energy and use it to solve for the maximum velocity:

$KE=\frac{1}{2}mv^2$

$7861.05J=\frac{1}{2}0.23kgv^2$

$2(7861.05J)=2(\frac{1}{2}0.23kgv^{2})$

$\frac{15722.1J}{0.23kg}{}=v^2$

$68356.96\frac{J}{kg}=v^{2}$

Since $J = \frac{kgm^{2}}{s^{2}}$ , that means that $\frac{J}{kg}=\frac{m^{2}}{s^{2}}$ .

$68356.96\frac{m^2}{s^2}=v^2$

$\sqrt{68356.96\frac{m^2}{s^2}}=\sqrt{v^2}$

$261.45\frac{m}{s}=v$

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Question

A spring with a spring constant of $325\frac{N}{m}$ has a mass of attached to one end. It is stretched a distance of . How much force is required to restore the spring to its equilibrium position?

Answer

The formula for the restoring force of a spring is:

$F=-k\Delta x$

Essentially, the restoring force is equal and opposite to the force required to stretch the spring. Note that the mass has no place in this calculation. We are given the spring constant and displacement, allowing us to calculate the force.

$F=-k\Delta x$

$F=-325\frac{N}{m}*0.64m$

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Question

A force is used to stretch a spring . What is the spring constant?

Answer

The formula for the force required to stretch or compress a spring is:

$F=k\Delta x$

We are given the force and the distance, allowing us to solve for the spring constant.

$\frac{41N}{0.11m}=k$

$372.73\frac{N}{m}=k$

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Question

What is the potential energy stored in a spring that is stretched and has a spring constant of $412\frac{N}{m}$ ?

Answer

Spring potential energy is given by the equation:

$PE=\frac{1}{2}kx^2$

We are given the spring constant and the distance that the spring is stretched. Using these values, we can find the energy stored in the spring.

$PE=(\frac{1}{2})(412\frac{N}{m})(0.67m)^2$

$PE=(206\frac{N}{m})(0.4489m^2)$

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Question

A spring with a spring constant of $180\frac{N}{m}$ is compressed . What is the potential energy in the spring?

Answer

The formula for potential energy in a spring is:

$PE=\frac{1}{2}kx^2$

We are given the values for the spring constant and displacement, allowing us to calculate the potential energy.

$PE=\frac{1}{2}(180\frac{N}{m})(0.22m)^2$

$PE=\frac{1}{2}(180\frac{N}{m})(0.0484m^2)$

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Question

How much force is required to compress a spring if it has a spring constant of $180\frac{N}{m}$ ?

Answer

The formula for compression force in a spring is:

We are given the value for the spring constant and the displacement, allowing us to solve for the force required.

$F=180\frac{N}{m}*0.22m$

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Question

A girl bounces on a massless pogo stick. If the spring constant for the stick is $15,000\frac{N}{m}$ , what is the maximum compression of the spring?

$g=-9.8\frac{m}{s^2}$

Answer

There are two forces at work here: the force due to gravity and the restoring force of the spring. We can set these two forces equal to one another because the forces must be in equilibrium when the spring is compressed at its maximum point.

Expand this equation by using the formulas for gravitational and spring force, respectively.

Plug in our given values for the girl's mass, gravitational acceleration, and the spring constant. Using these values, we can solve for the displacement of the spring.

$(36kg)(-9.8\frac{m}{s^2})=-(15,000\frac{N}{m})(x)$

$-352.8N=-15,000\frac{N}{m}*x$

$\frac{-352.8N}{-15,000\frac{N}{m}}=x$

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Question

A spring is stretched in the horizontal direction. If the spring requires of force to restore it to its original position, what is the spring constant?

Answer

To solve this problem, use Hooke's law.

We know the force of the spring and the distance it is displaced. Using these values, we can solve for the spring constant.

$\frac{-800N}{1.32m}=-k$

$-606.1\frac{N}{m}=-k$

$606.1\frac{N}{m}=k$

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Question

A spring with a spring constant of $1500\frac{N}{m}$ is compressed . A object is placed at the end of the compressed spring and the spring is released. What is the maximum velocity of the object?

Answer

For this problem, use the law of conservation of energy. Assuming no other forces are acting upon the object the initial spring potential energy will be equal to the maximum final kinetic energy.

$PE_{spring}=KE_f$

Expand this equality with the formulas for each type of energy.

$\frac{1}{2}kx^2=\frac{1}{2}mv^2$

We are given the spring constant and displacement, allowing us to complete the left side of the equation. We are also able to plug in the mass to the left side of the equation.

$\frac{1}{2}(1500\frac{N}{m})(0.87m)^2=\frac{1}{2}(0.67kg)v^2$

Solve to isolate and solve for the velocity variable.

$(750\frac{N}{m})(0.7569m^2)=(0.335kg)v^2$

$\frac{567.68J}{0.335kg}=v^2$

$1694.57\frac{m^2}{s^2}=v^2$

$\sqrt{1694.57\frac{m^2}{s^2}}=\sqrt{v^2}$

$41.17\frac{m}{s}=v$

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Question

How much potential energy is generated by a spring with a spring constant of $820\frac{N}{m}$ if it is stretched from equilibrium?

Answer

Spring potential energy is equal to half of the spring constant times the compression/stretching distance squared:

$PE=\frac{1}{2}k(\Delta x)^2$

Using the given values for the spring constant and displacement, we can solve for the energy.

$PE=\frac{1}{2}(820\frac{N}{m})(0.79m)^2$

$PE=(410\frac{N}{m})(0.6241m^2)$

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Question

A spring with a spring constant of $200 \frac{N}{m}$ is compressed . What is the potential energy stored in the spring?

Answer

The equation for spring potential energy is $PE=\frac{1}{2}kx^2$ .

Plug in the given values for the distance and spring constant to solve for the potential energy.

$PE=(\frac{1}{2}) (200\frac{N}{m}) (- 0.5m)^2$

Remember, since the spring was compressed, it has a negative displacement. The resultant potential energy will be positive as, when released, the displacement will be along the positive horizontal axis.

$PE=(\frac{1}{2})(200\frac{N}{m}) (0.25m^2)$

$PE=\frac{1}{2}(50\frac{N\cdot m^2}{m})$

$PE=\frac{1}{2} (50Nm)$

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