A block of mass 0.250 kg is placed on top of a light, vertical spring of force constant 5 000 N/m and pushed downward so that the spring is compressed by 0.100 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?

Explanation:

Given that,

Mass place on spring is

M=0.25kg

Force constant of spring is

K=5000N/m

Compression of spring is

e=0.1m

The mass starts from rest, then it's initial velocity u=0m/s

Maximum height?

Energy in spring is converted to change in Potential energy and change in kinetic energy

Energy in spring is given as

U=½ke²

U=½×5000×0.1²

U=25J

Gravitational potential energy is given as

P.E = mgh

Where

m is mass

g is gravitational constant 9.81m/s²

h is height reached by object

Change in potential energy is given as

Not the initial height h of the object is zero h=0, so we want to find the final height (H)

∆P.E= mgH - mgh

∆P.E= 0.25×9.81×H -0.25×9.81×0

∆P.E= 2.4525H -0

∆P.E= 2.4525H

Change in kinetic energy is 0J, the object starts from rest then, it initial velocity is 0m/s and when it get to maximum height, the velocity at maximum height is 0m/s,

Change in kinetic energy is given as

∆K.E=½mVf² - ½mVi²

∆K.E= 0J

Therefore,

U= ∆P.E +∆K.E

25=2.4525H +0

2.4525H=25

H=25/2.4525

H=10.194m

Maximum height reached by the ball is 10.194m

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.