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High School Physics : Using Spring Equations

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Example Question #11 : Harmonic Motion

A 41N force is used to stretch a spring 0.11m . What is the spring constant?

Possible Answers:

$0.496\frac{N}{m}$

$0.223\frac{N}{m}$

$372.73\frac{N}{m}$

$3388.43\frac{N}m{}$

$4.51\frac{N}{m}$

Correct answer:

$372.73\frac{N}{m}$

Explanation:

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.

41N=k*0.11m

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

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

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Example Question #11 : Using Spring Equations

What is the potential energy stored in a spring that is stretched 0.67m and has a spring constant of $412\frac{N}{m}$ ?

Possible Answers:

184.95J

138.02J

92.47J

276.04J

568.64J

Correct answer:

92.47J

Explanation:

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.

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

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

PE=92.47J

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Example Question #11 : Using Spring Equations

A spring with a spring constant of $325\frac{N}{m}$ has a mass of 0.22kg attached to one end. It is stretched a distance of {0.64m} . How much force is required to restore the spring to its equilibrium position?

Possible Answers:

-133.12N

-208N

-104N

-66.56N

-71.5N

Correct answer:

-208N

Explanation:

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.

$F=-k\Delta x$

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

F=-208N

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Example Question #221 : Motion And Mechanics

A spring with a spring constant of $180\frac{N}{m}$ is compressed 0.22m . What is the potential energy in the spring?

Possible Answers:

0.05J

9.21J

4.36J

19.8J

8.71J

Correct answer:

4.36J

Explanation:

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.

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

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

PE=4.36J

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Example Question #11 : Using Spring Equations

How much force is required to compress a spring 0.22m if it has a spring constant of $180\frac{N}{m}$ ?

Possible Answers:

219.6N

39.6N

818.18N

19.8N

4.36N

Correct answer:

39.6N

Explanation:

The formula for compression force in a spring is:

F=kx

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

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

F=39.6N

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Example Question #16 : Using Spring Equations

A spring is stretched 1.32m in the horizontal direction. If the spring requires -800N of force to restore it to its original position, what is the spring constant?

Possible Answers:

$606.1\frac{N}{m}$

$800\frac{N}{m}$

$356.4\frac{N}{m}$

$459.1\frac{N}{m}$

$1056\frac{N}{m}$

Correct answer:

$606.1\frac{N}{m}$

Explanation:

To solve this problem, use Hooke's law.

F=-kx

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

-800N=-(k)(1.32m)

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

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

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

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Example Question #11 : Using Spring Equations

A spring with a spring constant of $1500\frac{N}{m}$ is compressed 0.87m . A 0.67kg object is placed at the end of the compressed spring and the spring is released. What is the maximum velocity of the object?

Possible Answers:

$41.17\frac{m}{s}$

There is insufficient information to solve

$190.79\frac{m}{s}$

$1694.6\frac{m}{s}$

$13.79\frac{m}{s}$

Correct answer:

$41.17\frac{m}{s}$

Explanation:

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.

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

567.68J=(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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Example Question #11 : Using Spring Equations

How much potential energy is generated by a spring with a spring constant of $820\frac{N}{m}$ if it is stretched 0.79m from equilibrium?

Possible Answers:

209.8J

255.9J

647.8J

511.8J

295.5J

Correct answer:

255.9J

Explanation:

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.

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

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

PE=255.9J

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Example Question #1 : Forces

A 36kg 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?

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

Possible Answers:

42.52m

0.0004m

0.02m

0.14m

0.2m

Correct answer:

0.02m

Explanation:

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.

F_g=F_s

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

mg=-kx

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.

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

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

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

0.02m=x

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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 #11 : Harmonic Motion

Example Question #11 : Using Spring Equations

Example Question #11 : Using Spring Equations

Example Question #221 : Motion And Mechanics

Example Question #11 : Using Spring Equations

Example Question #16 : Using Spring Equations

Example Question #11 : Using Spring Equations

Example Question #11 : Using Spring Equations

Example Question #1 : Forces

All High School Physics Resources

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