Physics - Using Spring Equations

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 $181\frac{N}{m}$ ?

$261.45\frac{m}{s}$

$7861.05\frac{m}{s}$

$15722.1\frac{m}{s}$

$68356.96\frac{m}{s}$

$130.73\frac{m}{s}$

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 . 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:

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

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

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

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

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

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

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

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

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

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

$372.73\frac{N}{m}$

$4.51\frac{N}{m}$

$0.223\frac{N}{m}$

$0.496\frac{N}{m}$

$3388.43\frac{N}m{}$

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.

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

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

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?

$606.1\frac{N}{m}$

$1056\frac{N}{m}$

$459.1\frac{N}{m}$

$800\frac{N}{m}$

$356.4\frac{N}{m}$

Explanation

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.

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

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

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

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 $200\frac{N}{m}$ ?

$3.78\frac{m}{s}$

$14.29\frac{m}{s}$

$4\frac{m}{s}$

$8\frac{m}{s}$

$7.56\frac{m}{s}$

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 . 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:

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

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

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

, 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:

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

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

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

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

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

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

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

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

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

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.

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

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

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

Explanation

The formula for spring potential energy is:

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

Plug in our given values and solve:

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

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

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

, so:

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

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

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.

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

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

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?

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

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.

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.

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

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

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

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?

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.

$F=-k\Delta x$

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

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

What force is required to compress it ?

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.

$F=k\Delta x$

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

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.

Using Spring Equations

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation