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AP Physics B : Using Spring Equations

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Example Questions

Example Question #1 : Ap Physics 1

A spring is compressed 5 cm from equilibrium by a force of 100 N . What is the spring constant for this spring?

Possible Answers:

$5 \frac{N}{m}$

$20 \frac{N}{m}$

$500 \frac{N}{m}$

$2000 \frac{N}{m}$

$200 \frac{N}{m}$

Correct answer:

$2000 \frac{N}{m}$

Explanation:

We use Hooke's law equation to relate the force, displacement, and spring constant:

F_s=-kx

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

100N=-k(-0.05m)

Note that the displacement is negative, since the spring is compressed. For springs, compressions represents a negative displacement, while stretching represents a positive displacement.

$k=\frac{100N}{0.05m}=2000\frac{N}{m}$

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Example Question #21 : Newtonian Mechanics

A mass of weight 20N is suspended vertically from a spring and stretches it 10cm from equilibrium. What is the energy stored in the spring?

Possible Answers:

0.25J

0.5J

Correct answer:

Explanation:

First, we need to solve for the spring constant by using the force on the spring. We can use Hooke's Law:

F=-kx

The magnitude of the force on the spring will be equal to the force of gravity on the mass, which is given to be 20N . The distance the spring it stretched is in the downward direction, so we must use a negative displacement. Use these values to calculate the spring constant.

F=(20N)=-k(-0.10m)

$k=200\frac{N}{m}$

Next, use the spring energy equation with the displacement and spring constant to solve for the energy stored in the spring.

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

$E=\frac{1}{2}(200\frac{N}{m})(-0.10m)^2$

E=1J

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Example Question #1 : Other Potential Energy

A 4kg block is sliding on a horizontal frictionless floor at a speed of $2.5\frac{m}{s}$ and runs into a horizontal spring. The spring has a spring constant of $30\frac{N}{m}$ . What is the maximum compression of the spring after the collision?

Possible Answers:

0.83m

0.95m

0.91m

0.78m

0.87m

Correct answer:

0.91m

Explanation:

First, we calculate the kinetic energy of the block as it slides:

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

$KE=\frac{1}{2}(4kg)(2.5\frac{m}{s})^2=12.5J$

We know that all of the kinetic energy is converted to spring potential energy during the collision. We can use the equation for the potential energy of a spring and set it equal to the kinetic energy. Then, solve for the compression distance:

$KE=U_{spring}=\frac{1}{2}kx^2$

$12.5J=\frac{1}{2}(30\frac{N}{m})(x)^2$

$x^2=\frac{12.5J}{\frac{1}{2}(30\frac{N}{m})}=0.833m^2$

$x=\sqrt{0.833m^2}=0.91m$

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AP Physics B : Using Spring Equations

Study concepts, example questions & explanations for AP Physics B

All AP Physics B Resources

Example Questions

Example Question #1 : Ap Physics 1

Example Question #21 : Newtonian Mechanics

Example Question #1 : Other Potential Energy

All AP Physics B Resources

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