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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 $200 \frac{N}{m}$ is compressed 0.5m . What is the potential energy stored in the spring?

Possible Answers:

25J

100J

50J

12J

200J

Correct answer:

25J

Explanation:

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.

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

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

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

PE=25J

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Example Question #6 : Harmonic Motion

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

What force is required to compress it 0.1m ?

Possible Answers:

0.1N

-2000N

-2N

200N

-20N

Correct answer:

-20N

Explanation:

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.

$F=k\Delta x$

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

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 $16\frac{N}{m}$ .

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

Possible Answers:

1.28m

0.78m

3.33m

328m

Correct answer:

1.28m

Explanation:

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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Example Question #2 : Using Spring Equations

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

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

Possible Answers:

-1.11m

-17.92m

-2.82m

-0.45m

-14.13m

Correct answer:

-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 ( $PE_{spring}$ ).

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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Example Question #2 : Using Spring Equations

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

Possible Answers:

40N

100N

180N

200N

4000N

Correct answer:

40N

Explanation:

For this problem use Hooke's Law:

$F=k\Delta x$

Plug in our given values:

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

F=40N

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Example Question #4 : Using Spring Equations

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

Possible Answers:

1686.92N

843.46N

168.69N

7861.05N

190.32N

Correct answer:

1686.92N

Explanation:

For this problem use Hooke's Law:

$F=k\Delta x$

Plug in our given values:

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

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 , if it has a spring constant of $200\frac{N}{m}$ ?

Possible Answers:

40J

44J

0.04J

Correct answer:

Explanation:

The formula for spring potential energy is:

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

Plug in our given values and solve:

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

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

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

J = N*m , so:

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

PE=4J

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Example Question #6 : Using Spring Equations

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

Possible Answers:

7861.05J

8095.58J

4047.79J

1686.92J

3930.52J

Correct answer:

7861.05J

Explanation:

The formula for spring potential energy is

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

Plug in our given values and solve:

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

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

PE = 7861.05N*m

J = N*m , so:

PE=7861.05J

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Example Question #2 : Using Spring Equations

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

Possible Answers:

$4\frac{m}{s}$

$7.56\frac{m}{s}$

$3.78\frac{m}{s}$

$14.29\frac{m}{s}$

$8\frac{m}{s}$

Correct answer:

$3.78\frac{m}{s}$

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 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:

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

Plug in our given values and solve:

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

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

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

J = N*m , so:

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

PE=4J

The formula for kinetic energy is:

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

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

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

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

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

8J=0.56kg*v^2

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

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

Since $J = \frac{kg*m^{2}}{s^{2}}$ , 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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Example Question #4 : Using Spring Equations

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

Possible Answers:

$261.45\frac{m}{s}$

$7861.05\frac{m}{s}$

$68356.96\frac{m}{s}$

$15722.1\frac{m}{s}$

$130.73\frac{m}{s}$

Correct answer:

$261.45\frac{m}{s}$

Explanation:

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

We can use the law of conservation of energy to see 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:

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

Plug in our given values and solve:

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

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

PE=7861.05N*m

J = N*m , so:

PE=7861.05J

The formula for kinetic energy is:

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

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

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

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

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

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

15722.1J=0.23kg*v^2

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

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

Since $J = \frac{kg*m^{2}}{s^{2}}$ , 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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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 #6 : Harmonic Motion

Example Question #1 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #4 : Using Spring Equations

Example Question #3 : Using Spring Equations

Example Question #6 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #4 : Using Spring Equations

All High School Physics Resources

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