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High School Physics : Using Spring Equations

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Example Questions

Example Question #1 : Using Spring Equations

A spring with a spring constant of $200 \frac{N}{m}$ $\displaystyle 200 \frac{N}{m}$ is compressed 0.5m $\displaystyle 0.5m$. What is the potential energy stored in the spring?

Possible Answers:

25J $\displaystyle 25J$

200J $\displaystyle 200J$

12J $\displaystyle 12J$

100J $\displaystyle 100J$

50J $\displaystyle 50J$

Correct answer:

25J $\displaystyle 25J$

Explanation:

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

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

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

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.

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

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

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

PE=25J $\displaystyle PE=25J$

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Example Question #1 : Using Spring Equations

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

What force is required to compress it 0.1m $\displaystyle 0.1m$?

Possible Answers:

-2000N $\displaystyle -2000N$

-2N $\displaystyle -2N$

0.1N $\displaystyle 0.1N$

-20N $\displaystyle -20N$

200N $\displaystyle 200N$

Correct answer:

-20N $\displaystyle -20N$

Explanation:

For this problem, use Hooke's law:

$F=k\Delta x$ $\displaystyle F=k\Delta x$

In this formula, $\displaystyle k$ is the spring constant, $\Delta x$ $\displaystyle \Delta x$ is the compression of the spring, and $\displaystyle F$ 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.

$F=k\Delta x$ $\displaystyle F=k\Delta x$

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

F=-20N $\displaystyle F=-20N$

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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Example Question #1 : Using Spring Equations

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

If a force of 20.5N $\displaystyle 20.5N$ is used to stretch out the spring, what is the total displacement of the spring?

Possible Answers:

1.28m $\displaystyle 1.28m$

0.78m $\displaystyle 0.78m$

$\displaystyle 1m$

3.33m $\displaystyle 3.33m$

328m $\displaystyle 328m$

Correct answer:

1.28m $\displaystyle 1.28m$

Explanation:

For this problem, use Hooke's law:

$F=k\Delta x$ $\displaystyle F=k\Delta x$

In this formula, $\displaystyle k$ is the spring constant, $\Delta x$ $\displaystyle \Delta x$ is the compression of the spring, and $\displaystyle F$ 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.

$F=k\Delta x$ $\displaystyle F=k\Delta x$

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

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

$1.28m=\Delta x$ $\displaystyle 1.28m=\Delta x$

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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Example Question #2 : Using Spring Equations

A vertical spring with a spring constant of $11.1\frac{N}{m}$ $\displaystyle 11.1\frac{N}{m}$ is stationary. An 8kg $\displaystyle 8kg$ mass is attached to the end of the spring. What is the maximum displacement that the spring will stretch?

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

Possible Answers:

-14.13m $\displaystyle -14.13m$

-2.82m $\displaystyle -2.82m$

-1.11m $\displaystyle -1.11m$

-17.92m $\displaystyle -17.92m$

-0.45m $\displaystyle -0.45m$

Correct answer:

-14.13m $\displaystyle -14.13m$

Explanation:

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 ( $PE_{spring}$ $\displaystyle PE_{spring}$).

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:

$PE_{mass}=PE_{spring}$ $\displaystyle PE_{mass}=PE_{spring}$

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

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

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

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

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

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

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

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

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

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

$-14.13m=\Delta y$ $\displaystyle -14.13m=\Delta y$

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

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Example Question #1 : Using Spring Equations

How much force is required to compress a spring 0.2m $\displaystyle 0.2m$, if it has a spring constant of $200\frac{N}{m}$ $\displaystyle 200\frac{N}{m}$ ?

Possible Answers:

200N $\displaystyle 200N$

4000N $\displaystyle 4000N$

100N $\displaystyle 100N$

180N $\displaystyle 180N$

40N $\displaystyle 40N$

Correct answer:

40N $\displaystyle 40N$

Explanation:

For this problem use Hooke's Law:

$F=k\Delta x$ $\displaystyle F=k\Delta x$

Plug in our given values:

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

F=40N $\displaystyle F=40N$

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Example Question #2 : Using Spring Equations

How much force is required to compress a spring 9.32m $\displaystyle 9.32m$, if it has a spring constant of $181\frac{N}{m}$ $\displaystyle 181\frac{N}{m}$ ?

Possible Answers:

843.46N $\displaystyle 843.46N$

168.69N $\displaystyle 168.69N$

190.32N $\displaystyle 190.32N$

$\displaystyle 1686.92N$

$\displaystyle 7861.05N$

Correct answer:

$\displaystyle 1686.92N$

Explanation:

For this problem use Hooke's Law:

$F=k\Delta x$ $\displaystyle F=k\Delta x$

Plug in our given values:

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

$\displaystyle F=1686.92N$

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Example Question #3 : Using Spring Equations

How much potential energy is created by compressing a spring 0.2m $\displaystyle 0.2m$, if it has a spring constant of $200\frac{N}{m}$ $\displaystyle 200\frac{N}{m}$ ?

Possible Answers:

0.04J $\displaystyle 0.04J$

44J $\displaystyle 44J$

$\displaystyle 4J$

40J $\displaystyle 40J$

$\displaystyle 8J$

Correct answer:

$\displaystyle 4J$

Explanation:

The formula for spring potential energy is:

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

Plug in our given values and solve:

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

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

$PE = \frac{1}{2}*8N*m$ $\displaystyle PE = \frac{1}{2}*8N*m$

J = N*m $\displaystyle J = N*m$, so:

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

PE=4J $\displaystyle PE=4J$

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Example Question #3 : Using Spring Equations

How much potential energy is generated by compressing a spring 9.32m $\displaystyle 9.32m$, if it has a spring constant of $181\frac{N}{m}$ $\displaystyle 181\frac{N}{m}$?

Possible Answers:

$\displaystyle 1686.92J$

$\displaystyle 3930.52J$

$\displaystyle 8095.58J$

$\displaystyle 4047.79J$

$\displaystyle 7861.05J$

Correct answer:

$\displaystyle 7861.05J$

Explanation:

The formula for spring potential energy is

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

Plug in our given values and solve:

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

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

$\displaystyle PE = 7861.05N*m$

J = N*m $\displaystyle J = N*m$, so:

$\displaystyle PE=7861.05J$

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Example Question #2 : Using Spring Equations

A 0.56kg $\displaystyle 0.56kg$ mass is placed at the end of a spring. The spring is compressed 0.2m $\displaystyle 0.2m$. What is the maximum velocity of the mass if the spring has a spring constant of $200\frac{N}{m}$ $\displaystyle 200\frac{N}{m}$ ?

Possible Answers:

$7.56\frac{m}{s}$ $\displaystyle 7.56\frac{m}{s}$

$4\frac{m}{s}$ $\displaystyle 4\frac{m}{s}$

$3.78\frac{m}{s}$ $\displaystyle 3.78\frac{m}{s}$

$14.29\frac{m}{s}$ $\displaystyle 14.29\frac{m}{s}$

$8\frac{m}{s}$ $\displaystyle 8\frac{m}{s}$

Correct answer:

$3.78\frac{m}{s}$ $\displaystyle 3.78\frac{m}{s}$

Explanation:

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 $\displaystyle PE_i=KE_f$. 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:

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

Plug in our given values and solve:

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

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

$PE=\frac{1}{2}*8N*m$ $\displaystyle PE=\frac{1}{2}*8N*m$

J = N*m $\displaystyle J = N*m$, so:

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

PE=4J $\displaystyle PE=4J$

The formula for kinetic energy is:

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

Since $PE_{i}=KE_{f}$ $\displaystyle PE_{i}=KE_{f}$, that means that KE=4J $\displaystyle KE=4J$.

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

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

$4J=\frac{1}{2}*0.56kg*v^2$ $\displaystyle 4J=\frac{1}{2}*0.56kg*v^2$

$\displaystyle 8J=0.56kg*v^2$

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

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

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

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

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

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

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Example Question #13 : Harmonic Motion

A 0.23kg $\displaystyle 0.23kg$ mass is placed at the end of a spring. If the spring is compressed 9.32m $\displaystyle 9.32m$, what will be the mass's final velocity if the spring has a spring constant of $181\frac{N}{m}$ $\displaystyle 181\frac{N}{m}$?

Possible Answers:

$130.73\frac{m}{s}$ $\displaystyle 130.73\frac{m}{s}$

$15722.1\frac{m}{s}$ $\displaystyle 15722.1\frac{m}{s}$

$7861.05\frac{m}{s}$ $\displaystyle 7861.05\frac{m}{s}$

$68356.96\frac{m}{s}$ $\displaystyle 68356.96\frac{m}{s}$

$261.45\frac{m}{s}$ $\displaystyle 261.45\frac{m}{s}$

Correct answer:

$261.45\frac{m}{s}$ $\displaystyle 261.45\frac{m}{s}$

Explanation:

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

The formula for spring potential energy is:

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

Plug in our given values and solve:

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

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

$\displaystyle PE=7861.05N*m$

J = N*m $\displaystyle J = N*m$, so:

$\displaystyle PE=7861.05J$

The formula for kinetic energy is:

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

Since $PE_{i}=KE_{f}$ $\displaystyle PE_{i}=KE_{f}$, we know that $\displaystyle KE=7861.05J$.

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

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

$7861.05J=\frac{1}{2}*0.23kg*v^2$ $\displaystyle 7861.05J=\frac{1}{2}*0.23kg*v^2$

$2(7861.05J)=2(\frac{1}{2}*0.23kg*v^{2})$ $\displaystyle 2(7861.05J)=2(\frac{1}{2}*0.23kg*v^{2})$

$\displaystyle 15722.1J=0.23kg*v^2$

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

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

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

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

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

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

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High School Physics : Using Spring Equations

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Using Spring Equations

Example Question #1 : Using Spring Equations

Example Question #1 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #1 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #3 : Using Spring Equations

Example Question #3 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #13 : Harmonic Motion

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