Question

A spring that has a force constant of 1050 N/m is mounted vertically on the ground. A block of mass 1.40 kg is dropped from rest from height of 1.40 m above the free end of the spring. By what distance does the spring compress?

Answer 1

given that ::

spring force constant, k = 1050 N/m

mass of block, m = 1.4 kg

height, h = 1.4m

g = 9.8 m/s²

to find distance, y = ?

total height, H = h + y               { equation 1 }

using conservation of energy,

1 / 2 k y² = mgH

1 / 2 k y ² = mg (h + y)                     { from 1 }

1 / 2 k y ² = mgh + mgy

1 / 2 k y ² - mgy = mgh               { equation 2 }

This is a quadratic equation in y. Let's put it in the form ay² + by + c = 0 so that we can use the quadratic formula to find y.

inserting values in above equation,

1 / 2 k y ² - mgy - mgh = 0

1 / 2 (1050 N/m) y ² - (1.4 kg) (9.8 m/s²) y - (1.4 kg) (9.8 m/s²) (1.4m) = 0

525 y ² - 13.72 y - 19.208 = 0                     { equation 3 }

comparing with quadratic equation, ay² + by + c = 0

here, a = 525, b = - 13.72, c = - 19.208

y =   - b +_-                              ( eqaution 4 }

putting all these value in equation 4,

y = - (-13.72) +

y =   - (-13.72) +    / 1050

y =   - (-13.72) + 201.308 /1050 = 215.028 / 1050

y = 0.204 m or y = - 0.178 m