Physics question please help!?

Question

A vertical spring(massless) whose k=900 N/m is attached to a table and compressed 0.150m. (a) What speed can it give a .3 kg ball when released? (b) How high above its original position will the ball fly?

Answer 1

The force will put out .150 x 900 N/m = 135N

F= m x a
135N=.3kg x a
a = 450 m/sec^2

The ball is originally at rest just before you release the spring so Vo =0; we have a, we are given s=.15 as the force of the spring will push the ball until it is back to relaxed position.

V= Sg rt. (Vo^2 + 2 as)
= substituting and solving yields
V= 11.62 m/sec ANSWER TO A

after leaving the spring- the acceleration changes from spring force to gravity a= 9.81 m/sec^2. the ball will rise until it stops (and then begins to fall) so V=0 at the top of its flight.

distance, s= (V^2 - Vo^2)/2a
0^2 - 11.62^2/2 x 9.81= 6.9 meters ANSWER TO B

Answer 2

Conservation of energy is the correct approach. However, you need to take into account the change in gravitational potential energy as the spring is released.

The energy stored in the spring accounts for the total potential energy change from the compressed position. Hence, you determine the maximum height first. The velocity is then determined from the difference between the max height and the uncompressed position as a change in potential energy that can be then converted to kinetic energy

Answer 3

remember the conservation energy law, because there is not friction forces. In this case:

(1/2)kx^2 = (1/2)mv^2, then

v^2 = kx^2 / m= 67.5 (m/s)^2 then v = 8.21 m/s

Now, you can apply again the energy conservation law:

(1/2) mv^2 = mgh, then

h = (1/2)v^2 / g = 3.44 m


Please, check the numeric values.
Good luck!!