Question

If a block of mass 800g is attached to a horizontal spring with spring constant 100 N/m, how would its maximum speed change compare to a block of mass 200g attached?

Answer 1

We know that the total mechanical energy on a spring mass system is given by:

`(1)/(2)mv^2+(1)/(2)kx^2`

We also know that the mechanical energy is conserved, this means that the energy oscillates between the kinetic energy of the block and the potential energy stored in the spring.

We also know that the maximum velocity happens when the potential energy is zero; this happens when the spring is not compress. With this in mind we notice that the maximum potential energy and the maximum kinetic energy is the same.

Assume that the maximum potential energy happens when x=A. In this case we know that the spring is stretch an amount A, then the potential energy is:

`(1)/(2)(100)A^2`

Now, when the kinetic energy is maximum it has to have the same value then:

`(1)/(2)mv^2=(1)/(2)100A^2`

Solving for the velocity we have that:

`\begin{gathered} v^2=(1)/(m)100A^2 \\ v=\sqrt[]{(1)/(m)100A^2} \\ v=\frac{10A}{\sqrt[]{m}} \end{gathered}`

For the mass of 800 gr we have that the maximum velocity will be:

`v=\frac{10A}{\sqrt[]{.8}}`

while for the mass of 200 gr the maximum velocity will be:

`v^(\prime)=\frac{10A}{\sqrt[]{.2}}`

(we use a prime here to so we can distinguish between the two blocks).

To compare this velocities we make the quotient:

`\begin{gathered} (v)/(v^(\prime))=\frac{\frac{10A}{\sqrt[]{.8}}}{\frac{10A}{\sqrt[]{.2}}} \\ (v)/(v^(\prime))=\frac{\sqrt[]{.2}}{\sqrt[]{.8}} \\ (v)/(v^(\prime))=\sqrt[]{(.2)/(.8)} \\ (v)/(v^(\prime))=\sqrt[]{(1)/(4)} \\ (v)/(v^(\prime))=(1)/(2) \end{gathered}`

This means that:

`v=(1)/(2)v^(\prime)`

Therefore the maximum speed of the 800 gr mass is half the maximum velocity of the 200 gr mass.