Question

A 1.25-kg block is on a horizontal surface with muk = 0.180, and is in contact with a lightweight spring with a spring constant of 529 N/m which is compressed. Upon release, the spring does 3.20 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed.

What is the velocity of the mass as it breaks contact with the spring?

Answer 1

Hi there!

We can find the compressed distance by using the equation for Spring Potential Energy:

`U = (1)/(2)kx^2`

U = Potential Energy (3.2 J)

k = Spring Constant (529 N/m)

x = Compression distance (? m)

Rearrange and solve for 'x'.

`x^2 = (2U)/(k)\\\\x = \sqrt{(2U)/(k)}\\\\x = \sqrt{(2(3.2))/(529)} = .11 m`

Now, we can find the velocity using the Work-Energy theorem. There is work done by friction in this instance, however.

`E_i = E_f\\`

Initially, there is only Spring Potential Energy. Then, some of that energy is converted to kinetic energy, while some is lost to friction.

Thus:

`U = K + W_f\\\\(1)/(2)kx^2 = (1)/(2)mv^2 + \mu mgd`

Rearrange to solve for velocity.

`(1)/(2)kx^2 - \mu mgd = (1)/(2)mv^2\\\\kx^2 - 2\mu mgd = mv^2\\\\v^2 = (kx^2 - 2\mu mgd)/(m)\\\\v = \sqrt{(kx^2 - 2\mu mgd)/(m)}`

Plug in the values and solve. (d = compression distance)

`v = \sqrt{(529(.11^2) - 2(.18)(1.25)(9.8)(0.11))/(1.25)} = \boxed{2.175 (m)/(s)}`