Question

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 14.9 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 3h0 above the spring, what would be the maximum compression of the spring?

Answer 1

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Step-by-step explanation:

a) W by the block on spring is given by

W=
`(1)/(2)`kx² =
`(1)/(2)`(530)(0.149)² = 5.88 J

b) Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg
`h_o` =
`(1)/(2)`kx² - mgx

And,
`h_o`= (
`(1)/(2)`kx² - mgx )/(mg) =
`[(1)/(2) (530)(0.149^2)-(0.645)(9.8)(0.149)`]/(0.645x9.8)

`h_o`= 0.78m

d) Now, if the initial initial height of block is 3
`h_o`

`h_o` = 3 x 0.78 = 2.34m

then,
`(1)/(2)`kx² - mgx - mg
`h_o` =0

`(1)/(2)`(530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum compression of the spring)