Ablock of mass 0.244 kg is placed on top of a light, vertical spring of force constant 5 075 n/m and pushed downward so that the spring is compressed by 0.106 m. after the block is released from rest, it travels upward and then leaves the spring. to what maximum height above the point of release does it rise? (round your answer to two decimal places.)

Answer with Explanations:

Given:

mass of block, m = 0.244 kg

spring constant, k = 5075 N/m

compression distance in spring, x = 0.106 m

Find how high mass reaches after release (measured from point of release)

Solution:

Use conservation of energy, assuming a light spring (of negligible mass)

stored spring potential energy = gravitational potential energy of block

PE1 = PE2

PE1 = kx^2/2 = 5075*0.106^2/2 = 28.512 J

PE2 = mgh = 0.244 * 9.8 * h = 2.3912 h J

Solve for h = 28.512 / 2.3912 = 11.923 m

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student